Let us call the angles x, y , and z. We know that these three angles must add up to 180; therefore,
[tex]x+y+z=180[/tex]If x is the smallest angle, we know that one of the angles (call it y) is 3 times this angle; therefore,
[tex]y=3x[/tex]And the other angle z is 20 more than the smallest angle x; therefore, we have
[tex]z=20+x[/tex]Putting the last two equations into the very first equation gives us
[tex]x+3x+20+x=180[/tex][tex]5x+20=180[/tex][tex]\textcolor{#FF7968}{\therefore x=32}[/tex]which is the measure of the smallest angle.
The other two angles are
[tex]y=3x=3(32)[/tex][tex]\textcolor{#FF7968}{y=96}[/tex]and
[tex]z=20+x=20+32[/tex][tex]\textcolor{#FF7968}{z=52.}[/tex]Hence, the measure of the smallest angle is 32,
The measure of the middle angle is 52
And the measure of the largest angle is 96.
-what is the measure of arc AB-what is the length of arc AB-what is the area of the shaded section-what is the area of the unshaded section
Given Data:
The radius of the circle is, r = 4
The angle is, 98.
The measure of angle that an arc makes at the center of the circle of which it is a part. Therefore, the measure of the arc AB is 90.
The length of the arc AB can be calculated as,
[tex]\begin{gathered} L=\frac{98}{360}\times2\times\pi\times r \\ \text{ =}\frac{98}{360}\times2\times3.14\times4 \\ \text{ =}6.84 \end{gathered}[/tex]The area of the shaded section is equal to the area of the arc, which can be calculated as,
[tex]\begin{gathered} A=\frac{98}{360}\times\pi\times r^2 \\ \text{ =}\frac{98}{360}\times3.14\times4^2 \\ \text{ =}13.67 \end{gathered}[/tex]The area of the unshaded region can be calculated by subtracting the area of the arc from the area of the total circle. The area of the total circle is,
[tex]A^{\prime}=\pi\times r^2=3.14\times4^2=50.24[/tex]Therefore the area of the unshaded region can be calculated as,
[tex]A=50.24-13.67=36.57[/tex]8 A marine biologist is using an underwater drone to study a delicatecoral reef. The linear equation 20y – 30x = -900 gives the drone'selevation, y, in meters from the surface of the water after x seconds.Graph the equation. What are the slope and y-intercept of the line?What part of the graph represents this situation?
Draw the equation 20y - 30x = -900 on the graph.
Simplify the equation to obtain the slope and intercept.
[tex]\begin{gathered} 20y-30x=-900 \\ 20y=30x-900 \\ y=\frac{30}{20}x-\frac{900}{20} \\ =\frac{3}{2}x-45 \end{gathered}[/tex]The equation for the slope of that y-intercept of the line will be equal to y = 3/2x - 45.
What is an equation?Equations are mathematical expressions that have two algebras on either side of an equal (=) sign. The expressions on the left and right are shown to be equal, demonstrating this relationship. L.H.S. = R.H.S. (left-hand side = right side) is a fundamental simple equation.
As per the given equation in the question, draw the graph.
The given equation is 20y - 30x = -900.
Solving the equation to get the slope,
20y - 30x = -900
20y = -900 + 30x
y = -900/20 + 30x/20
y = (3/2)x - 45
To know more about equation:
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which expression is a factor of 9 - 4.17 F 3-1 G 1 9-1 There are no real factors
hello
to solve this question, we simply need to identify the factors of the equation and then check which of them is in the options
[tex]9r^2-4r+1=0[/tex]this equation has no real factors
the answer to this question is option J
ABC is a triangle for which < ABC = 90°, AB= 10cm and BC=15cm, Find Ac? how do I solve this?
The Pythagorean theorem states:
[tex]c^2=a^2+b^2[/tex]where a and b are the legs and c is the hypotenuse of a right triangle.
Applying this theorem to triangle ABC (see the above diagram):
[tex]AC^2=AB^2+BC^2[/tex]Substituting with AB = 10 cm, BC = 15 cm, and solving for AC:
[tex]\begin{gathered} AC^2=10^2+15^2 \\ AC^2=100+225 \\ AC^2=325 \\ AC=\sqrt[]{325} \\ AC\approx18\operatorname{cm} \end{gathered}[/tex]1) For the following pairs equations explain why the equations are equivalent (or not) -11(x-2)=8 x-2=8+11 2)For the following pairs of equations explain why the equations are equivalent. (Or Not!) -3(2x+9)=12 2x+9=-4
hello
the first equation given was
[tex]-11(x-2)=8x-2=8+11[/tex]let's resolve each side of the equations
for the first one,
[tex]\begin{gathered} -11(x-2)=8x-2 \\ -11x+22=8x-2 \\ \text{collect like terms} \\ 8x+11x=22+2 \\ 19x=24 \\ \text{divide both sides by coefficient of x} \\ \frac{19x}{19}=\frac{24}{19} \\ x=\frac{24}{19} \end{gathered}[/tex]now let's test for the other side of the equation
[tex]\begin{gathered} 8x-2=8+11 \\ 8x-2=19 \\ 8x=19+2 \\ 8x=21 \\ \text{divide both sides by the coeffiecient of x} \\ \frac{8x}{8}=\frac{21}{8} \\ x=\frac{21}{8} \end{gathered}[/tex]from the calculations above, the two equations are not equal
[tex]\frac{24}{19}\ne\frac{21}{8}[/tex]What is the solution to the following equation? 3(X - 4)- 5 = X - 3 O A. X = 12 B. X = 8 O C. x=7 O D. *= 3
The initial equation is:
[tex]3(x-4)-5=x-3[/tex]Now we can distribute the 3 into the parenthesis so:
[tex]3x-12-5=x-3[/tex]Now we move all variable to the left of the equation and the constants to the right so:
[tex]3x-x=-3+12+5[/tex]Finally we simplify so:
[tex]\begin{gathered} 2x=14 \\ x=\frac{14}{2} \\ x=7 \end{gathered}[/tex]So the answer is C) x = 7
f(x) = 5x2 + 2x − 3What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph.
EXPLANATION:
Given;
We are given a quadratic function as shown below;
[tex]f(x)=5x^2+2x-3[/tex]Required;
We are required to show the steps required to graph this function.
Step-by-step solution;
To do this, we would need to set up a table of values for x and y.
We shall take input values that is x, and use this to get corresponding output value, that is y.
[tex]\begin{gathered} When\text{ }x=0: \\ y=5(0)^2+2(0)-3 \end{gathered}[/tex][tex]y=0+0-3=-3[/tex]Therefore, we have the coordinates (0, -3)
[tex]\begin{gathered} When\text{ }x=1: \\ y=5(1)^2+2(1)-3 \end{gathered}[/tex][tex]y=5+2-3=4[/tex]This gives us the coordinates, (1, 4)
[tex]\begin{gathered} When\text{ }x=2: \\ y=5(2)^2+2(2)-3 \end{gathered}[/tex][tex]y=20+4-3=18[/tex]This too gives us the coordinates, (2, 18).
Note that the same can be done for the left side of the graph (where the x values are negative).
Hence, using the same procedure as shown above, we would have;
[tex]\begin{gathered} When\text{ }x=-1,y=0 \\ That\text{ }is,\text{ }(-1,0) \end{gathered}[/tex][tex]\begin{gathered} When\text{ }x=-2,y=13 \\ That\text{ }is,\text{ }(-2,13) \end{gathered}[/tex][tex]\begin{gathered} When\text{ }x=-3,y=36 \\ That\text{ }is,\text{ }(-3,36) \end{gathered}[/tex]We can go on and plot as many points as the graph page can accommodate using this same procedure.
Next step is to connect the points as shown by the coordinates.
Using a graphing calculator, the graph of the quadratic function given will be as shown below;
Sixty students were asked to state their favourite subject chosen from their school timetable the table below was obtained. How can I draw a bar chart to show the information given. How can I draw a pie chart to represent the information?
Okay, here we have this:
Considering the provided information, we are going to draw a bar chart and a pie chart to show the information given, so we obtain the following:
Bar chart:
For this case we will assign a bar to each subject and on the vertical axis it will indicate the number of people who prefer the subject, then we have:
Pie chart:
For this case then we will draw a circle, and each part of the circle will represent the percentage of students who prefer that subject, so for example we can divide it into 60 parts and for each subject color the number of parts that prefer that subject of a color, it will be something So:
Then, for each sector, the number of students who prefer it is first observed, followed by the percentage that this subject represents.
Question 9 (3 points)y = -x3 + 2x + 3d: {-2, 1,2} r: {3Blank 1:Blank 2:IBlank 3:
The graph of a quadratic function with vertex (3,2) is shown in the figure below. Write the domain and range using interval notation.
The domain of a function is the set of possible x-values that the function can take, and the range is the set of possible y-values that the function will have.
As can be seen in the graph, as the function increase to the left the x-values will approach to -infinite and as it increases to the right, the x-values will approach to +infinite, then the domain is:
[tex](-\infty,\infty)[/tex]The vertex of the function is located at (3,2) then, the y-values start at 2, and increase infinitely, then the range is:
[tex]\lbrack2,\infty)[/tex]I will show you the pic
we have the expression
[tex]\frac{4.8\cdot10^8}{1.2\cdot10^4}\cdot2.2\cdot10^{(-6)}[/tex]Solve the quotient first
so
[tex]\begin{gathered} \frac{4.8\cdot10^8}{1.2\cdot10^4}=\frac{4.8}{1.2}\cdot10^{(8-4)} \\ \\ 4\cdot10^4 \end{gathered}[/tex]substitute in the original expression
[tex]\begin{gathered} 4\cdot10^4\cdot2.2\cdot10^{(-6)}=(4\cdot2.2)\cdot10^{(4-6)} \\ 8.8\cdot10^{(-2)} \end{gathered}[/tex]the answer is8.8x10^-2Find the value of 9y+1 given that -2y-1=5.Simplify your answer as much as possible.
Answer
The value of 9y + 1 is -26
Explanation:
Given the below equation
-2y - 1 = 5
Step 1: find y
To find y, firstly collect the like terms
-2y = 5 + 1
-2y = 6
Divide both sides by -2
-2y / -2 = 6/-2
y = -3
Given 9y + 1
Substitute the value of y = -3
9(-3) + 1
= -27 + 1
= - 26
Therefore, the value of 9y + 1 is -26
Question 8 of 10 Multiply (x - 2)(3x + 4) using the distributive property. Select the answer choice showing the correct distribution. A. (x-2)(3x) + (x-2)(4) B. (X)(3x + 4) + (x - 2) C. (x-2)(3x) + (3x)(4) D. (X)(3x) + 4(x) + 3x + 4 SUBMIT
To multiply
[tex]\begin{gathered} (x-2)\times(3x+4) \\ (x-2)\times3x+(x-2)(4) \end{gathered}[/tex]Hence, option A is correct.
simplify (2^x-6)(^x-2)
solution
[tex]\begin{gathered} (2\sqrt[]{x}-6)(\sqrt[]{x}-2) \\ 2\sqrt{x}\sqrt{x}+2\sqrt{x}\mleft(-2\mright)+\mleft(-6\mright)\sqrt{x}+\mleft(-6\mright)\mleft(-2\mright) \\ 2\sqrt{x}\sqrt{x}-2\cdot\: 2\sqrt{x}-6\sqrt{x}+6\cdot\: 2 \\ 2x-10\sqrt{x}+12 \end{gathered}[/tex]answer is:
[tex]2x-10\sqrt[]{x}+12[/tex]The population of mice doubles every year. There are 10 mice in the population. Write an exponential function to model this situation:Determine the population of the mice in 11 years. How many years will it take for the population of mice to reach at least 10 million?
Let's list down the given information in the problem.
1. initial value = 10 mice
2. rate = doubles every year = 200%
3. time = 11 years
4. final value = at least `10 million
The equation of an exponential function goes by this pattern:
[tex]f(x)=ab^x[/tex]where a = initial value, b = growth rate, x = time in years
From the given information, we can write an exponential model of the situation by plugging in those given data (1 and 2 only) to the pattern above.
[tex]f(x)=10(2)^x[/tex]The exponential model is f(x) = 10(2)^x as shown above.
After 11 years, the population will be: (plug in x = 11 to the model)
[tex]\begin{gathered} f(x)=10(2)^{11} \\ f(x)=10(2048) \\ f(x)=20,480 \end{gathered}[/tex]After 11 years, the population of the mice will have been 20, 480.
To calculate how many years it will take the population to reach at least 10 million, we will have to assume that f(x) = 10 million and solve for x.
[tex]10,000,000=10(2)^x[/tex][tex]\begin{gathered} \text{Divide both sides by 10.} \\ 1,000,000=2^x \\ Convert\text{ to logarithmic form.} \\ \log _21,000,000=x \\ x\approx19.93 \\ x\approx20 \end{gathered}[/tex]Thus, it will take approximately 20 years for the population of the mice to at least reach 10 million.
Plato classes The pH scale measures the acidity of a liquid as a function of its hydrogen ion (H+) concentration. How does the H+ concentration of a solution with a pH of 2 compare with that with a pH of 1?
Let H₁ be the concentration of a solution with a pH of 1, and H₂ be the concentration of a solution with a pH of 2, then:
[tex]\begin{gathered} 1=-\log(H_1), \\ 2=-\log(H_2). \end{gathered}[/tex]Then:
[tex]\begin{gathered} H_1=10^{-1}=0.1, \\ H_2=10^{-2}=0.01. \end{gathered}[/tex]Therefore:
[tex]H_1=0.1H_2.[/tex]Therefore the H concentration in a solution with a pH of 2 is 0.1 times of a solution with a pH of 1.
Answer: Second option.
Which expression best estimates 6O 7-2○ 6-18-3○ 3-2NEATRY10:DAITEELKATEER EEN ANDER HEAADEMPIRENTE] PAR MEpar spenna#] [3]5\y\£x{] =>DEEEN TOEGERHENDEROgeJENTERE3위 - 10
Step 1
Given;
[tex]6\frac{3}{4}\div1\frac{2}{3}[/tex]Required; To find Which expression best estimates the expression.
Step 2
[tex]\begin{gathered} 6\frac{3}{4}\text{ =6.75}\approx7 \\ 1\frac{2}{3}\approx2 \end{gathered}[/tex]Thus the answer will be;
[tex]7\div2[/tex]At 12:30 pm Jane made a telephone call which cost 60 pence. If she had waited until after 1:00 pm she could have had 2 minutes longer for the same cost. Calls after 1:00 pm are 1 penny per minute cheaper than calls before 1:00 pm. Let x minutes be the length of the call and n pence be the cost per minute at 12:30. i Express n in terms of x.ii Show that x + 2x - 120 = 0. iii Solve this equation and hence find the length of Jane's call.
Let x be the length of the call per minute;
Let n be the cost per minute (in pence);
[tex]\begin{gathered} \text{ cost = (cost/minute)}\times length \\ \end{gathered}[/tex]That is, at 12:30pm;
[tex]\begin{gathered} 60=n\times x \\ 60=nx \\ n=\frac{60}{x} \end{gathered}[/tex]Similarly at 1:00pm;
[tex]\begin{gathered} 60=(n-1)\times(x+2) \\ n-1=\frac{60}{x+2} \\ n=\frac{60}{x+2}+1 \end{gathered}[/tex]Then, we have expressed n in terms of x at 12:30pm and 1:00pm, then we have;
[tex]n=\frac{60}{x+2}+1=\frac{60}{x}[/tex]Simplifying further, we have;
[tex]\begin{gathered} \frac{60}{x+2}+1=\frac{60}{x} \\ M\text{ultiply through by x(x+2);} \\ 60x+x^2+2x=(x+2)60 \\ 60x+x^2+2x=60x+120 \\ 60x-60x+x^2+2x-120=0 \\ x^2+2x-120=0 \end{gathered}[/tex]Thus, the equation is correct.
(iii) By simplifying the equation above, we have;
[tex]\begin{gathered} x^2+2x-120=0 \\ x^2+12x-10x-120=0 \\ x(x+12)-10(x+12)=0 \\ x-10=0\text{ or x+12=0} \\ x=10\text{ or x = -12} \end{gathered}[/tex]Thus, the length of Jane's call is 10minutes
For csc 330:a) state value of the ratio exactlyb) find one equivalent expressionc) draw a diagram to illustarte.
For csc 330:
a) state value of the ratio exactly
b) find one equivalent expression
c) draw a diagram to illustarte.
SolutionPart A[tex]-2[/tex]Part bRecall:
[tex]csc=\frac{1}{sin330}[/tex]Part c9√x +7√x simplify the expression
We will have the following:
[tex]9\sqrt{x}+7\sqrt{x}=16\sqrt{x}[/tex]What is the least common multiple of 62 +39 - 21 and 6x² +54x+84? O 6x² +54x+84 6x² +93x +63 62³ +52x² + 111x − 42 O 12x³ + 102x² + 114 - 84
SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given expressions
[tex]6x^2+39x-21\text{ }and\text{ }6x^2+54x+84[/tex]STEP 2: Define the least common multiple
The Least Common Multiple ( LCM ) is also referred to as the Lowest Common Multiple ( LCM ) and Least Common Divisor ( LCD) . For two integers a and b, denoted LCM(a,b), the LCM is the smallest positive integer that is evenly divisible by both a and b.
STEP 3: Find the LCM
Factorize the first expression
[tex]6x^2+39x-21=3(2x-1)(x-7)[/tex]Factorize the second expression:
[tex]6x^2+54x+84=3(x+2)(x+7)[/tex]Calculating the LCM, we have:
[tex]\begin{gathered} \mathrm{Multiply\:each\:factor\:with\:the\:highest\:power:} \\ 2\cdot \left(2x-1\right)\cdot \:3\cdot \left(x+2\right)\cdot \left(x+7\right) \\ Simplify \\ 6\left(2x-1\right)\left(x+2\right)\left(x+7\right) \end{gathered}[/tex]Evlauating the result gives:
[tex]12x^3+102x^2+114x-84[/tex]Hence, the LCM is:
[tex]12x^3+102x^2+114x-84[/tex]A sphere and a cylinder each have the same radius. The cylinder has a height that is triple the radius. Which figure has a greater volume, the cylinder or the sphere?
we calculate the volume of the two solids
Sphere
[tex]V_s=\frac{4}{3}\times\pi\times r^3[/tex]Cylinder
[tex]\begin{gathered} V_c=\pi\times r^2\times h \\ V_c=\pi\times r^2\times3r \\ V_c=\pi\times3r^3 \end{gathered}[/tex][tex]\begin{gathered} \frac{4}{3}\times\pi\times r^3 \\ \\ 3\times\pi\times r^3 \end{gathered}[/tex]the comparison allows us to see that the volume of the cylinder is greater than that of the sphere
Sketch the asymptotes and the graph of the function. Identify the domain and range. y = 7/x+4 -7Choose the correct graph below (the asymptotes are shown as red dashed lines).
In the horizontal asymptotes:
If the exponent of the numerator is the same as the exponent of the denominator, then:
[tex]\begin{gathered} y=\text{ }\frac{a}{b} \\ y=\text{ }\frac{7}{1} \\ y=7 \end{gathered}[/tex]But, when there's a translation, then y= k, which means y= - 7.
In the vertical asymptotes:
You have to equal the denominator to 0, so:
[tex]\begin{gathered} x+\text{ 4= 0 } \\ x=\text{ -4} \end{gathered}[/tex]The domain is all real numbers except -4 and the range is all real numbers except -7.
Now we have to check the graphs.
The correct graph is C. When x is -4 and y is -7.
Find the surface area of the cylinder to the nereast tenth of a square unit. Use 3.14 for PI.
Given:
Height of the cylinder is, h = 18.2 cm.
Radius of the cylinder is, r = 3 cm.
The objective is to find the surface area of the cylinder.
The formula to find the surface area of the cylinder is,
[tex]SA=2\pi r^2+2\pi rh[/tex]Now substitute the given values in the above equation.
[tex]\begin{gathered} SA=(2\cdot3.14\cdot3^2)+(2\cdot3.14\cdot3\cdot18.2) \\ =(2\cdot3.14\cdot9)+(2\cdot3.14\cdot3\cdot18.2) \\ =56.5+342.9 \\ =399.4cm^2 \end{gathered}[/tex]Hence, option (B) is the correct answer.
Indicate the answer choice that best completes the statement or answers the question. ALGEBRA Find the value of x in each figure. 2) 65° O a. 23 O b. 25 O c. 63 O d. 113
They are complementary angles, so,
65 + (x + 2) = 90°
Solve for x
x + 2 = 90 - 65
x + 2 = 25
x = 25 - 2
x = 23°
Result x = 23
b) The procedure is the same, they are complementary angles.
43 + (x - 7) = 90°
Solve for x
x - 7 = 90 - 43
x - 7 = 47
x = 47 + 7
x = 54 This is the result Letter C.
Write the sum using sigma notation: 123 +4+***+122= sum l ^ A B where
GIVEN
The sum is given to be:
[tex]1+2+3+4+...+122=\sum_{n=1}^AB[/tex]SOLUTION
The sigma notation for writing sums can be described as shown in the image below:
From the given information, each successive term of the sum increases by 1. This means that the formula for the terms will be:
[tex]\Rightarrow n[/tex]The last value of the sum is 122.
Therefore, the sigma notation will be:
[tex]\Rightarrow\sum_{n=1}^{122}\:n[/tex]Hence, the answers are:
[tex]\begin{gathered} A=122 \\ B=n \end{gathered}[/tex]Use the following function rule to find (2). f(x) = (-5 - 2x)? ? f(2) =
Using the function f(x) = -5 - 2x, in order to evalute the value of f(2), we just need to use the value of x = 2 and calculate the value of the function. So we have that:
[tex]\begin{gathered} f(x)=-5-2x \\ f(2)=-5-2\cdot2 \\ f(2)=-5-4 \\ f(2)=-9 \end{gathered}[/tex]So the value of f(2) is -9.
Find a dishes between an order pairs (-2 -1) and (-5,-4) round your solution to the nearest 10th, if necessary.
To solve this problem, we will use the following formula for the distance between two points (x₁,y₁) and (x₂,y₂):
[tex]d=\sqrt[]{(x_1-x_2)^2+(y_1-y_2)^2}.[/tex]Substituting the given points in the above formula, we get:
[tex]d=\sqrt[]{(-2-(-5))^2+(-1-(-4))^2}.[/tex]Simplifying the above result, we get:
[tex]d=\sqrt[]{(-2+5)^2+(-1+4)^2}=\sqrt[]{3^2+3^2}=\sqrt[]{18}.[/tex]Therefore:
[tex]d\approx4.2.[/tex]Answer:
[tex]4.2[/tex]How do you find the common difference of an arithmetic sequence? Ex. 28,18,8,-2
Answer:
To find the common difference of an arithmetic sequence, we compute the difference of an element of the sequence and its predecessor.
Example: Given the sequence 28, 18, 8, -2, the common difference is:
[tex]\begin{gathered} 18-28=-10, \\ 8-18=-10, \\ -2-8=-10. \end{gathered}[/tex]The function f(x) = x³ - 3x² + 2x rises as x grows very large. O A. True O B. False
We have to find what happens with f(x) when x grows very large.
The function is:
[tex]f(x)=x^3-3x^2+2x[/tex]When x grows very large, the term that has a higher degree will be the one that have the greatest effect in the value of f(x).
In this function, the term with the highest degree is x³.
As x grows very large, so does x³. Then, we can conclude that f(x) will also rise as does x.
Answer: True.