In the figure, AC is perpendicular to DB. Which of tgehe following statements is TRUE?
Answers
Complementary angles are two angles with a sum of 90 degrees and for supplementary angles, the sum is 180 degrees.
From the figure we can infer that ,
1 + 2 + 3 = 90
Option A.
Related Questions
Given triangle FGH is similar to triangle LMN, which must be true?
Answers
Options A, B, and E
Explanations:See below the diagramatic representations of the two congruent triangles
Since △FGH ∼ △LMN:
All corresponding sides are similar
The ratio of two corresponding sides are equal
Corresponding angles are congruent
Therefore:
[tex]\begin{gathered} \frac{FG}{LM}=\text{ }\frac{FH}{LN} \\ FH\text{ }\sim LN \\ m7. Identify a horizontal or vertical stretch or compression of the function f(x)=√x by observing the equation of the function g(x)=√3x.
Answers
Given:
The function is:
[tex]\begin{gathered} f(x)=\sqrt{x} \\ \\ g(x)=\sqrt{3x} \end{gathered}[/tex]Find-:
The horizontal or vertical stretch or compression of the function
Explanation-:
The graph of a function is:
The f(x) is:
[tex]f(x)=\sqrt{x}[/tex]The graph of function is:
[tex]g(x)=\sqrt{3x}[/tex]So, the both function is:
So, the graph is vertically stretch by √3
[tex]\begin{gathered} f(x)=\sqrt{x} \\ \\ g(x)=\sqrt{3x} \\ \\ g(x)=\sqrt{3}\sqrt{x} \end{gathered}[/tex]
Suppose you chose to solve the following system of equations by multiplying the first equation by 2 then by what number would you multiply the second equation to eliminate the variable by adding?
Answers
Answer:
6
Explanation:
Given the below system of equations;
[tex]\begin{gathered} 4x-3y=6\ldots\ldots\ldots\text{Equation 1} \\ 6x+y=10\ldots\ldots\ldots\text{Equation 2} \end{gathered}[/tex]If we multiply the first equation by 2, we'll have;
[tex]8x-6y=12\ldots\ldots\text{.Equation 3}[/tex]To be able to eliminate a variable by adding, we can notice that if multiply Equation 2 by 6. If we do, we'll have;
[tex]36x+6y=60\ldots\ldots\ldots\text{Equation 4}[/tex]If we add Equation 3 and 4, we'll have;
[tex]\begin{gathered} (36+8)x+(-6+6)y=60+12 \\ 44x+0y=72 \\ 44x=72 \end{gathered}[/tex]So if we multiply the second equation by 6, we will be able to eliminate a variable as seen above.
Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.An employee at a construction company is ordering interior doors for some new houses that are being built. There are 5 one-story houses and 5 two-story houses on the west side of the street, which require a total of 115 doors. On the east side, there are 2 one-story houses and 5 two-story houses, which require a total of 85 doors. Assuming that the floor plans for the one-story houses are identical and so are the two-story houses, how many doors does each type of house have?
Answers
"There are 5 one-story houses and 5 two-story houses on the west side of the street, which require a total of 115 doors." translates to 5x + 5y = 115
"there are 2 one-story houses and 5 two-story houses, which require a total of 85 doors." translates to 2x + 5y = 85
[tex]\begin{gathered} \text{Solve by elimination} \\ 5x+5y=115 \\ 2x+5y=85 \\ \text{Multiply the second equation by -1, then add the equations together.} \\ (5x+5y=115) \\ -1(2x+5y=85 \\ \downarrow \\ 5x+5y=115 \\ -2x-5y=-85 \\ \text{Add these equations to eliminate y:} \\ 3x=30 \\ \text{Divide both sides by 3} \\ \frac{3x}{3}=\frac{30}{3} \\ x=10 \\ \text{Substitute }x=10\text{ to any of the equation} \\ 5x+5y=115 \\ 5(10)+5y=115 \\ 50+5y=115 \\ 5y=115-50 \\ 5y=65 \\ \frac{5y}{5}=\frac{65}{5} \\ y=13 \end{gathered}[/tex]Substitute x = 10 for one story houses, and y = 13 for two story houses.
[tex]\begin{gathered} \text{West Side Houses} \\ 5x+5y=115 \\ 5(10)=50 \\ 5(13)=65 \\ \text{There are 50 doors for one-story house and 65 for the two-story house on west side} \\ \\ \text{East Side houses} \\ 2x+5y=85 \\ 2(10)=20 \\ 5(13)=65 \\ \text{There are 20 doors for one-story house and 65 for the two-story house on the east side} \end{gathered}[/tex]8x-2y=343x+6y=33 solve the system using the elimination method if there is no solution put no solution if there is infinite solutions put infinite
Answers
8x - 2y = 34
3x + 6y = 33
Multiply the first equation by 3, and add both equations to eliminate y
24x -6y = 102
+
3x + 6y = 33
__________
27x = 135
x= 135/27
x= 5
Replace x=5 on any equation and solve for y
8(5) - 2y = 34
40 - 2y= 34
-2y= 34 - 40
-2y= -6
y= -6/-2
y=3
Solution:
x= 5
y= 3
Cost for copies is a linear function of the number of copies. If 125 copies cost $29.00, and 175 copies cost $37.00, write a formula for copy cost as a linear function of the number of copies. Then find how much it would cost tomake 700 copies
Answers
ANSWER
f(x) = 0.16x + 9
Cost to make 700 copies f(700)= $121
Let y=f(x) be the cost of copies and let x be the number of copies
( 125, 29) and (175, 37)
x₁=125 y₁=29 x₂=175 y₂=37
We first need to find the slope(m) using the formula below:
[tex]\text{slope}(m)=\frac{37-29}{175-125}[/tex][tex]=\frac{8}{50}[/tex][tex]=0.16[/tex]Next, is to find the intercept (b)
Substitute m=0.16 x=125 and y=29 into y=mx+b and solve for b
29 = 0.16(125) + b
29 = 20 + b
29 - 20 = b
b = 9
The linear function can be formed by substituting m=0.16 and b=9 into the formula y = mx+b
f(x) = 0.16x + 9
To find the cost of 700 copies, substitute x=700 into the above and evaluate.
f(700) = 0.16(700) + 9 = 112 + 9 = $121
Can u help me answer question 25 BTW this is lesson 6 the distributive property Question: Mrs.singh brought 9 folders and 9 notebooks. the cost of each folder was $2 50. each notebook cost $4. write two equivalent expressions and then find the total cost
Answers
∵ The number of the folders = 9
∵ The cost of each folder = $2.50
∴ The cost of the folders = 9 x 2.50
what are the angles of rotation for a rectangle ?
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A rectange has four sides. Two sides with the same length and other two sides of the same length.
This means that if you rotate the rectangle you will obtain the same figure only for rotation of 180°.
You can notice the previous conclusion in the following image:
The rectangle with black lines is the original one, and the rectangle with red lines is the rotated rectangle with a rotation of 180°. You can observe that for such a rtation yo obtain the same figure.
A diagram of a rectangular pool with a diagonal of 50 meters is shown below. Connected to the pool is a square shaped kid’s pool.What is the area of the kid’s pool in the diagram? Ulysses says the area of the kid’s pool is 900 square meters.Ursula says the area of the kid’s pool is 120 square meters. Which student is correct?
Answers
Let's find the length of the side of the pool.
Let it be x, so the diagram is:
Using the pythagorean theorem, we can figure out x. Shown below:
[tex]\begin{gathered} x^2+40^2=50^2 \\ x^2=50^2-40^2 \\ x^2=900 \\ x=\sqrt[]{900} \\ x=30 \end{gathered}[/tex]So, 30 meters is the side length of the square (pool).
The area of a square is:
[tex]A=s^2[/tex]Where "s" is the side length
We found s to be 30, thus the area of the pool is:
[tex]\begin{gathered} A=s^2 \\ A=(30)^2 \\ A=900 \end{gathered}[/tex]Thus,
Ulysses is correct.Can you please help me with 24Write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equation of asymptotes
Answers
24)
Given:
[tex]-9x^2+72x+16y^2+16y+4=0[/tex]We need to find the hyperbola in standard form
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]The given equation can be written as follows.
[tex]-9(x^2+8x)+16(y^2+y)+4=0[/tex][tex]-9(x^2+2\times4x+4^2-4^2)+16(y^2+2\times\frac{1}{2}y+\frac{1}{2^2}-\frac{1}{2^2})+4=0[/tex][tex]-9(x^2+2\times4x+4^2)-(-9)4^2+16(y^2+2\times\frac{1}{2}y+\frac{1}{2^2})-(16)\frac{1}{2^2}+4=0[/tex][tex]-9(x^2+2\times4x+4^2)+144+16(y^2+2\times\frac{1}{2}y+\frac{1}{2^2})-4+4=0[/tex][tex]-9(x^2+2\times4x+4^2)+144+16(y^2+2\times\frac{1}{2}y+\frac{1}{2^2})=0[/tex][tex]\text{ Use (a+b)}^2=a^2+2ab+b^2\text{.}[/tex][tex]-9(x+4)^2+144+16(y+\frac{1}{2})^2=0[/tex]Subtracting 144 from both sides, we get
[tex]-9(x+4)^2+144+16(y+\frac{1}{2})^2-144=0-144[/tex][tex]-9(x+4)^2+16(y+\frac{1}{2})^2=-144[/tex]Dividing both sides by (-144), we get
[tex]\frac{-9\mleft(x+4\mright)^2}{-144}+\frac{16(y+\frac{1}{2})^2}{-144}=-\frac{144}{-144}[/tex][tex]\frac{\mleft(x+4\mright)^2}{16}-\frac{(y+\frac{1}{2})^2}{9}=1[/tex][tex]\frac{\mleft(x+4\mright)^2}{4^2}-\frac{(y+\frac{1}{2})^2}{3^2}=1[/tex]Hence we get the standard form of the hyperbola.
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]Comapring with standerd form, we get
[tex]h=-4,k=-\frac{1}{2},a=4\text{ and b=3.}[/tex]The coordinates of the vertices are
[tex](h+a,k)\text{ and }(h-a,k)\text{ }[/tex]Substitute h=-4, a=4 and k=-1/2.
[tex](-4+4,-\frac{1}{2})\text{ and }(-4-4,-\frac{1}{2})[/tex][tex](0,-\frac{1}{2})\text{ and }(-8,-\frac{1}{2})[/tex][tex](0,-0.5)\text{ and }(-8,-0.5)[/tex]Hence the vertices are (0, -0.5) and (-8, -0.5).
The distance between foci is 2c,
[tex]c^2=a^2+b^2[/tex]Substitute a=4 and b=3 to find the value of c.
[tex]c^2=4^2+3^2[/tex][tex]c^2=16+9=25^{}[/tex][tex]c=\pm\sqrt[]{25}[/tex][tex]c=\pm5[/tex]The measure of distance can not be negative.
[tex]c=5[/tex]The coordinates of the foci are
[tex](h+c,k)\text{ and }(h-c,k)[/tex]Substitute h=-4, k=-1/2 and c=5, we get
[tex](-4+5,-\frac{1}{2})\text{ and }(-4-5,-\frac{1}{2})[/tex][tex](1,-\frac{1}{2})\text{ and }(-9,-\frac{1}{2})[/tex][tex](1,-0.5)\text{ and }(-9,-0.5)[/tex]
Hence the foci are
[tex](1,-0.5)\text{ and }(-9,-0.5)[/tex]The equation of the asymptotes are
[tex]y=\frac{b}{a}(x-h)+k\text{ and }y=-\frac{b}{a}(x-h)+k[/tex]Substitute a=4,b=3,h=-4 and k=-1/2, we get
[tex]y=\frac{3}{4}(x-(-4))-\frac{1}{2}\text{ and }y=-\frac{3}{4}(x-(-4))-\frac{1}{2}[/tex][tex]y=\frac{3}{4}(x+4)-\frac{1}{2}\text{ and }y=-\frac{3}{4}(x+4)-\frac{1}{2}[/tex][tex]y=\frac{3}{4}x+\frac{3}{4}\times4-\frac{1}{2}\text{ and }y=-\frac{3}{4}x+(-\frac{3}{4})\times4-\frac{1}{2}[/tex][tex]y=\frac{3}{4}x+3-\frac{1}{2}\text{ and }y=-\frac{3}{4}x-3-\frac{1}{2}[/tex][tex]y=\frac{3}{4}x+3\times\frac{2}{2}-\frac{1}{2}\text{ and }y=-\frac{3}{4}x-3\times\frac{2}{2}-\frac{1}{2}[/tex][tex]y=\frac{3}{4}x+\frac{6-1}{2}\text{ and }y=-\frac{3}{4}x-\frac{(6+1)}{2}[/tex][tex]y=\frac{3}{4}x+\frac{5}{2}\text{ and }y=-\frac{3}{4}x-\frac{7}{2}[/tex][tex]y=0.75x+2.5\text{ and }y=-0.75x-3.5[/tex]Hence the equations of the asymptotes are
[tex]y=0.75x+2.5\text{ and }y=-0.75x-3.5[/tex]Results:
The equation for the hyperbola in standard form is
[tex]\frac{\mleft(x+4\mright)^2}{4^2}-\frac{(y+\frac{1}{2})^2}{3^2}=1[/tex]The vertices are
[tex](0,-0.5)\text{ and }(-8,-0.5)[/tex]The foci are
[tex](1,-0.5)\text{ and }(-9,-0.5)[/tex]The equation of asymptotes are
[tex]y=0.75x+2.5\text{ and }y=-0.75x-3.5[/tex]Use cross products to determine whether the proportion below is true or false.420459Is the proportion true or false?A. False. The numerators of the two ratios are not equal.B. False. The denominators of the two ratios are not equal.O C. False. The cross products are not equal.O D. True. The cross products are equal.ba
Answers
We want to find the cross product of:
[tex]\frac{4}{9}\text{ and }\frac{20}{45}[/tex]Then, we have a kind of X:
We multiply the numbers inside the red oval:
4 x 45 = 180
and the blue oval
9 x 20 = 180
Then, the cross products are equal and the proportion
[tex]\frac{4}{9}=\frac{20}{45}\text{ is true}[/tex]Answer: D. The cross products are equalLet E be the event that a corn crop has an infestation of earworms, and let B be the event that a corn crop has an investigation of corn borers.
Answers
We will see how to determine independent probabilities of two events with the help of set notations.
We will define two events that can occur independently as follows:
[tex]\begin{gathered} \text{Event ( E ) = Corn crop infected by earth worms} \\ \text{Event ( B ) = Corn crop infected by corn borers} \end{gathered}[/tex]The probability of occurrence of each event is defined by the prefix " p " preceeding each event notation as follows:
[tex]\begin{gathered} p\text{ ( E ) = 0.18} \\ p\text{ ( B ) = 0.11} \\ \text{p ( E \& B ) = 0.06} \end{gathered}[/tex]We can express the two events ( E and B ) as two sets. We will Venn diagram to express the events as follows:
We have expressed the two events E and B as circles which are intersecting. The region of intersection is the common between both events ( E & B ).
The required probability is defined as an instance when the corn crop is subjected to either earth worm or corn borers or both!
If we consider Venn diagram above we can see the required region constitutes of region defined by event ( E ) and event ( B ). We can sum up the regions defined by each event!
However, if we consider region E and B as stand alone we see that the common region is added twice in the algebraic sum of induvidual region E and B. We will discount the intersection region once to prevent over-counting. Therefore,
[tex]p\text{ ( E U B ) = p ( E ) + p ( B ) - p ( E \& B )}[/tex]The above rule is the also denoted as the rule of independent events ( probabilities ).
We will use the above rule to determine the required probability p ( E U B ) as follows:
[tex]\begin{gathered} p\text{ ( E U B ) = 0.18 + 0.11 - 0.06} \\ p\text{ ( E U B ) = 0.23} \end{gathered}[/tex]Therefore, the required probability is:
[tex]0.23\ldots\text{ Option D}[/tex]1b. Check if each of these points is a solution to the inequality 2y - X> 1:*SolutionNot a solution(0,2)OO(8, 1/2)OO(-7-3)O
Answers
Solution:
Given:
[tex]2y-x>1[/tex]To test the points that is a solution to the inequality;
[tex]\begin{gathered} At\text{ (0,2)} \\ x=0,y=2 \\ \\ \text{Substituting into the inequality,} \\ 2y-x>1 \\ 2(2)-0 \\ 4 \\ \text{Since 4>1, then the point (0,2) satisfies the inequality and is a solution} \end{gathered}[/tex]
Therefore, (0,2) is a solution.
[tex]\begin{gathered} At\text{ (8,}\frac{1}{2}\text{)} \\ x=8,y=\frac{1}{2} \\ \\ \text{Substituting into the inequality,} \\ 2y-x>1 \\ 2(\frac{1}{2})-8 \\ 1-8 \\ -7 \\ \\ \text{Since -7< 1, then the point (8,}\frac{1}{2}\text{) does not satisfy the inequality and is not a solution} \end{gathered}[/tex]
Therefore, (8, 1/2) is not a solution.
[tex]\begin{gathered} At\text{ (-7,-3)} \\ x=-7,y=-3 \\ \\ \text{Substituting into the inequality,} \\ 2y-x>1 \\ 2(-3)-(-7) \\ -6+7 \\ 1 \\ \\ \text{Since 1=1, then the point (-7,-3) does not satisfy the inequality and is not a solution} \end{gathered}[/tex]
Therefore, (-7,-3) is not a solution.
Also, the graph of the inequality is shown below with the given points.
It can be seen that only the point (0,2) falls within the region of the solution.
Therefore, the answer is summarized below;
The 60 members of a school’s glee club need to raise at least $5,000 for a trip to a national competition. The school agreed to contribute $1,000 toward the trip. Which inequality below shows the amount of money that each glee club member needs to raise for the trip?
Answers
Since the club needs to raise $ 5, 000
And the school agrees to contribute $1000
Then if the least amount each of the 60 members of the club is represented by x
then
[tex]1000\text{ + 60x }\leq\text{ 5000}[/tex]determine the smallest integer value of x in the solution of the following inequlaity.
Answers
we have
-3x-7 < 14
Solve for x
Adds 7 both sides
-3x < 14+7
-3x < 21
Divide by -3 both sides
Remember that
when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol
so
x >-7the solution is all real numbers greater than -7
{x}^{4} - 17x ^{2} + 16factor completely
Answers
We can make a u-substitution to make it into a quadratic.
Let
[tex]u=x^2[/tex]Performing this substitution, we have:
[tex]\begin{gathered} x^4-17x^2+16 \\ =(x^2)^2-17(x^2)+16 \\ \text{Making u-substitution,} \\ u^2-17u+16 \\ (u-16)(u-1) \end{gathered}[/tex]Now, taking it back to "x", we have:
[tex]\begin{gathered} (u-16)(u-1) \\ (x^2-16)(x^2-1) \end{gathered}[/tex]We can further break it down by using the rule:
[tex]a^2-b^2=(a+b)(a-b)[/tex]So, let's factor it out completely:
[tex]\begin{gathered} (x^2-16)(x^2-1) \\ ((x)^2-(4)^2)((x)^2-(1)^2) \\ (x-4)(x+4)(x-1)(x+1) \end{gathered}[/tex]Thus, the fully factored form is:
[tex](x-4)(x+4)(x-1)(x+1)[/tex]BAIf m/BAC = 90°, and m
Answers
ANSWER
m∠BAD = 22°
EXPLANATION
By the angle addition theorem, the sum of the measures of angles BAD and DAC is equal to the measure of angle BAC,
[tex]m\angle BAD+m\angle DAC=m\angle BAC[/tex]Replace the known values,
[tex]m\angle BAD+68\degree=90\degree[/tex]Solving for m∠BAD,
[tex]m∠BAD=90\degree-68\degree=22\degree[/tex]Hence, the measure of angle BAD is 22°.
The number of men and women receiving bachelors degrees each year has been steadily increasing. For the years 1970 through the projection of 2014, the number of men receiving degrees ( in thousands ) is given by the equation y=3.7+441, and for women, the equation is y=14.4+314, where x is the number of years after 1970 A. Use the substitution method to solve the system of equations Sketch a graph of the system of equations, write a sentence describing the trends for men and women receiving bachelor degrees
Answers
Answer:
a. x = 12
b. 12 years after 1970, the number of men graduating equals the number of women graduating.
c.
Explanation:
We have the system
[tex]\begin{gathered} y=3.7x+441 \\ y=14.4x+314 \end{gathered}[/tex]and we solve it by substituting the value of y from the first equation into the second equation:
[tex]14.4x+314=3.7x+441[/tex]subtracting 314 from both sides gives
[tex]14.4x=3.7x+127[/tex]subtracting 3.7 from both sides gives
[tex]10.7x=127[/tex]Finally, dividing both sides by 127 gives
[tex]x=\frac{127}{10.7}[/tex][tex]\boxed{x=11.869\ldots}[/tex]which rounded to the nearest whole number is
[tex]\boxed{x=12}[/tex]The solution tells us that 12 years after 1970, the number of men graduating equals the number of women graduating.
The graph of the equations is given below:
By looking at the slope we see that more women graduate than men and after 12 years, the number of graduating women is greater than men.
A 228-inch is cut into two pieces. One piece is five times the length of the other. Find the lengths of the two pieces. The short piece is ___ inches long.
Answers
Given:
A 228-inch is cut into two pieces.
One piece is five times the length of the other.
Required:
To find the length of the two pieces.
Explanation:
Let assign a value to the shorter piece. "x" as the length.
The other piece is 5 times that, or 5x.
Add them up and you get 6x.
[tex]\begin{gathered} x+5x=228 \\ 6x=228 \\ x=\frac{228}{6} \\ x=38inches \\ 5x=5\times38 \\ =190inches \end{gathered}[/tex]Final Answer:
The length of the two pieces are 38inches and 190inches.
The current (I) in a wire varies directly as the voltage (v) and inversely as the resistance (r). If the current is 27.5 amps when the voltage is 110 volts and the resistance is 4 ohms, find the current when the voltage is 125 volts and the resistance is 16 ohms. (Round your answer to two decimal places.)
Answers
ANSWER
Current (l) = 7.81 amps (rounded to 2 decimal places)
EXPLANATION
Declaration of Variables
Let l represent the current in a wire,
v represent the voltage, and
r represent the resistance
Desired Outcome
The current (l)
Equation formation
[tex]\begin{gathered} l\propto\frac{v}{r} \\ l\text{ = }\frac{kv}{r} \end{gathered}[/tex]where k is the constant of proportionality.
Determine the value of k given l = 27.5, v = 110, and r = 4.
[tex]\begin{gathered} l\text{ = }\frac{kv}{r} \\ 27.5\text{ = }\frac{k\times110}{4} \\ 110k\text{ = 27.5}\times4 \\ 110k\text{ = 110} \\ k\text{ = }\frac{110}{110} \\ k\text{ = 1} \end{gathered}[/tex]Find the current (l) given v = 125, and r = 16.
[tex]\begin{gathered} l\text{ = }\frac{kv}{r} \\ l\text{ = }\frac{1\times125}{16} \\ l\text{ = 7.8125 amps} \end{gathered}[/tex]Hence, the current (l) when the voltage is 125 volts and the resistance is 16 ohms is 7.81 amps (rounded to 2 decimal places).
Hello somebody else help me with the first part I need help with the other part.So I need help with:-8x=__x=__—+y=12Andy=__
Answers
a. The given equation is:
[tex]-8x+36=20[/tex]Subtract 36 from both sides:
[tex]\begin{gathered} -8x+36-36=20-36 \\ -8x=-16 \end{gathered}[/tex]Divide both sides by -8:
[tex]\begin{gathered} -8x/-8=-16/-8 \\ x=2 \end{gathered}[/tex]b. The given equation is:
[tex]x+y=12[/tex]Replace the x-value we find in the previous part, and solve:
[tex]2+y=12[/tex]Subtract 2 from both sides:
[tex]\begin{gathered} 2+y-2=12-2 \\ y=10 \end{gathered}[/tex]c. The answer is: (2,10)
What is the common difference? An=5+-3(n-1)
Answers
Answer
Common difference = -3
Common difference = -
Explanation
The nth term of an arithmetic progression is given as
aₙ = a + d (n - 1)
aₙ = nth term
a = First term
d = common difference
n = number of terms
Comparing this to the given general term
An = 5 + -3 (n - 1)
We see that
d = -3
For the second expression,
An = 5 + -8 (n - 1)
Comparing this with the general form,
aₙ = a + d (n - 1)
a = 5
d = -8
Common difference = -8
Hope this Helps!!!
The function f(x) = 22 + 1 is graphed below: x Is this a growth or decay function? growth Domain: Range: > y-intercept:
Answers
A function showing an exponential growth is modelled in the form,
y = Aoe^kt
where k is the rate of growth
If k is negative, then it is the rate of decay and the function would be a decay function
a. what is the the slope of the line through the points 5, 2 and 5, -3B. describe the line through the points given
Answers
a). the the slope of the line through the points 5, 2 and 5, -3
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{-3-2}{5-5} \\ m=-\frac{5}{0} \\ m=\infty \end{gathered}[/tex]So the slope is:
[tex]\begin{gathered} \tan \theta=m=\infty \\ \theta=90^{\circ} \end{gathered}[/tex]the expression -25t + 1250 represents the volume of liquid of a container after t seconds. the expression 50t + 250 represents the volume of liquid of another container after t seconds. what does the equation -25 + 1250= 50t +250 mean in this situation?
Answers
The volume of liquid (V1) in first container at any time 't' is given as,
[tex]V_1=-25t+1250[/tex]The volume of liquid (V2) in other container at any time 't' is given as,
[tex]V_2=50t+250[/tex]Consider the equation,
[tex]\begin{gathered} -25+1250=50t+250 \\ V_1=V_2 \end{gathered}[/tex]Thus, the given equation represents the situation when the volume in both the containers are equal.
Write an equation for a polynomial with all the following features: The degree is 5, the polynomial has 4 terms and the graph of the polynomial crosses the vertical axis at y=12
Answers
we must apply the formula
[tex]y=x^2+bx\times c[/tex][tex]3.25x^5+4x^{4^{}}\text{ }+5.5x^{3\text{ }}+x^2[/tex]now we apply the formula to find x,
[tex]\begin{gathered} h=\frac{b}{2(a)} \\ h=\frac{-4}{2(1)} \\ h=-2 \end{gathered}[/tex]having already x now we apply the formula to find the vertex at y that intersects at point 12.
[tex]\begin{gathered} f(-2)=3.25(-2)^{5\text{ }}+4(-2)^{4^{}}+5.5(-2)^{3^{}}\text{ }+(-2)^{2^{}} \\ f(-2)=104-64-44-8 \\ =-12 \end{gathered}[/tex]Could you please answer these and show me how to get to the answers?
Answers
Given the function:
[tex]y=x^2-4x-5[/tex]- Factor to find the x-intercept:
Factor and equal to zero for Vertexeach factor.
[tex]x^2-4x-5=(x-5)(x+1)[/tex]So:
[tex]\begin{gathered} x-5=0 \\ x=5 \\ \text{and} \\ x+1=0 \\ x=-1 \end{gathered}[/tex]a) The vertex of a parabola is:
[tex]x=-\frac{b}{2a}[/tex]The parameters of the parabola are:
a = 1, b = -4 and c = -5
[tex]x=-\frac{-4}{2(1)}=\frac{4}{2}=2[/tex]We find y:
[tex]y=(2)^2-4(2)-5=4-8-5=-9[/tex]Therefore the vertex of the parabola is: (2, -9)
b) x-intercepts are: (5,0) and (-1,0)
c) Vertex: (2, -9)
d) Graph:
- Table:
x y
-1 0
0 -5
1 -8
2 -9
3 -8
4 -5
5 0
6 7
Ken has a food truck that sells for 5 dollars and up. Bryan buys $20 worth of food. How much money will Ken make in one hour if the price of the food goes up by $10 every 2 minutes and 30 seconds for the tourist visits
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Let:
M(t) = Money earned by ken as a function of time
Let's asume ken earned 20 dollars for the purchase of Bryan
And let's make a conversion:
2 min * 60 s/1min = 120s
So, the food goes up every 150s
Hence the function M(t) would be:
Let's define t2= 150s
M(t) = $20 + ($5+$10t/t2)
Because he sells his food for $5 and the price goes up by $10 every 150s
Ok, now an hour has 3600s:
So:
3600s/150s = 24
We can conclude that in one hour the food went up in price 24 times
Now, using the function:
M(t) = $20 + ($5+$10t/t2)
For t = 3600
M(3600) = $20 + ($5 + $10 * (3600)/(150))
M(3600) = $20 + ($5 + $10*24)
M(3600) = $20 + $5 + $240 = $265
Ken will make $265 in one hour
please help , i don’t know what to do !
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Given the inequality:
[tex]v<-7[/tex]So, we will select the values of v that is less than (-7)
The solution on the number line is as shown in the following figure:
So, we will select the numbers that are shown in the following figure:
The answer is the numbers with the red color.
If the height of a gable end of a roof is 7.5 m and the rafters are 10.2 long, at what angle do the rafters slope, and how wide is the gable end at the base?
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Solution:
Consider the following diagram of the problem situation:
We need to find the angle alpha and the base b. To find the angle alpha, we can apply the following trigonometric identity:
[tex]\sin (\alpha)=\frac{h}{l}[/tex]Replacing the data of the problem in the above equation, we get:
[tex]\sin (\alpha)=\frac{7.5}{10.2}[/tex]now, applying the inverse function of the sine function, we get:
[tex]\alpha=\sin ^{-1}(\frac{7.5}{10.2})=\text{ 47.33}[/tex]then, the angle would be:
[tex]\alpha=47.33^{\circ}[/tex]On the other hand, to find the lenght of the base, we can apply the Pythagorean Theorem:
[tex]b\text{ = }\sqrt[]{l^2-h^2}\text{ = }\sqrt[]{(10.2)^2-(7.5)^2}\text{ = 6.91}[/tex]so that, we can conclude that the base measures:
[tex]b\text{ = 6.91}[/tex]