we will start with tan^2 theta - sin^2 theta
[tex]\tan ^2\text{ }\theta-sin^2\theta[/tex]
remember tan theta = sin theta / cos theta
[tex](\frac{\sin \theta}{\cos \theta})^2\text{ - }\sin ^2\theta[/tex][tex](\frac{\sin\theta}{\cos\theta})^2\text{ - }\sin ^2\theta\text{ }\ast\frac{\cos ^2\theta}{\cos ^2\theta}[/tex][tex]\frac{\sin ^2\theta(1-\cos ^2\theta)}{\cos ^2\theta}[/tex][tex]\tan ^2\theta\text{ (1 - }\cos ^2\theta\text{)}[/tex][tex]\tan ^2\theta\ast\sin ^2\theta[/tex]which of the following rules should be used to find the areas of a triangle
The area A of a triangle with base b and altitude h, as shown in the diagram:
Is given by the equation:
[tex]A=\frac{1}{2}b\times h[/tex]Therefore, the correct choice is:
Multiply 1/2 base by altitude.
Find b so that the vectors v=i−bj and w=9i−7j are orthogonal.
Explanation:
For those vectors to be orthogonal, v*w=0.
In this case:
[tex]\begin{bmatrix}{1} & {} \\ -{b} & {}\end{bmatrix}*\begin{bmatrix}{9} & {} \\ -{7} & {}\end{bmatrix}=1*9+-b*-7=0[/tex][tex]\begin{gathered} 9+7b=0 \\ 7b=-9 \\ b=-\frac{9}{7} \\ \\ \end{gathered}[/tex]Lets check:
If b=-9/7, then v=i-(-9/7)j, and we have:
[tex]v=i+\frac{9}{7}j[/tex]And multiplying,
[tex]\begin{bmatrix}{1} & {} \\ {\frac{9}{7}} & {}\end{bmatrix}*\begin{bmatrix}{9} & {} \\ {-7} & {}\end{bmatrix}=1*9+\left(-7*\frac{9}{7}\right?=9+-9=0[/tex]v and w are orthogonals.
Answer:
b=-9/7
Hi I am stuck on the last part of this question. Please assist. Thank you
Okay, here we have this:
Considering the provided graph and the information previously found in the previous parts of the question, we are going to determine the function that represents the graph shown, then we have:
Then to find the requested function we will substitute in the following formula:
[tex]y=Asin(Bx)+D[/tex]Where each letter represents:
A: Amplitude
B=2pi/period
D: Midline
Replacing:
[tex]y=2sin(\frac{2\ast\pi\ast x}{7})-2[/tex]Graph each leg of Kiley's trip with the time (in hours) along the x-axis and distance (in miles) on the y-axis. Complete questions part A - LPart A: the starting coordinates of the graph will be (0,0) or the origin. This point represents Kileys house. From here, if you drew a line segment that represents Kiley's car ride from her house to the airport, at what point would the line segment end? Car ride from Kiley's house to the airport 24.3 miles in 0.5 hour = 48.6
Kiley's house is our reference. The airport is at 24.3 miles of distance, and it took 0.5 hours to reach there. So, on the y-axis, you have to find 24.3 and on the x-axis, you have to find 0.5, and then connect that point with Kile's house, which is at Origin.
Translate each graph as specified below.(a) The graph of y = f(x) is shown. Translate it to get the graph of y = f(x+5).(b) The graph of y = g (x) is shown. Translate it to get the graph of y = g (x) + 2.
Given:
The graph is:
[tex]\begin{gathered} y=f(x) \\ \\ y=g(x) \end{gathered}[/tex]Find-:
(a) Translate to get the graph.
[tex]y=f(x+5)[/tex](b)
Translate to get a graph
[tex]y=g(x)+2[/tex]Explanation-:
The function is:
[tex]\begin{gathered} y=f(x) \\ \\ y=f(x+5) \end{gathered}[/tex]The graph is a shift of 5 units left on the x-axis.
That means the graph move 5 units on the negative x side.
(b)
[tex]y=g(x)[/tex]The function is translated is:
[tex]y=g(x)+2[/tex]The graph shifts 2 units up side iny-axis.
So, The graph
simplify to a single power of 3 .. 3^5 / 3
The given expression is
3^5/3
To simplify to a single power of 3, we would apply one of the rules of exponents which is expressed as
a^b/a^c = a^(b - c)
By comparing the given expression with this rule,
a = 3
b = 5
c = 1 because 3 is the same as 3^1
By applying the rule, it becomes
3^(5 - 1)
= 3^4
Identify the function that relects f(x)=5x^3-3 across the x-axis and shifts it 2 units up.
To find the new function, follow the steps below.
Step 01: Reflect the function over the x-axis.
After reflecting the function f(x) over the x-axis, the new function is -f(x).
Then:
[tex]\begin{gathered} f(x)=5x^3-3 \\ -f(x)=-(5x^3-3) \\ -f(x)=-5x^3+3 \end{gathered}[/tex]Step 02: Shift the function 2 units up.
When the function is shifted n units up, the new function if f(x) + n.
The,
[tex]\begin{gathered} h(x)=-f(x)+2 \\ h(x)=-5x^3+3+2 \\ h(x)=-5x^3+5 \end{gathered}[/tex]Answer:
[tex]h(x)=-5x^3+5[/tex]
One carat equals how much atomic mass units? One milligram equals how many atomic mass units? The possible answers are: a) 1.2 x 10^-25b) 1.2 x 10^ 23c)6.02 x 10^-22d)6.02 x 10^ 20
We are given the following units of measure;
[tex]1\text{ Atomic mass}=8.3\times10^{-24}Carat[/tex]To determine the equivalent as 1 carat, we would divide the quantity of 1 atomic mass in carat by the value in carat.
This is shown below;
[tex]\begin{gathered} 1\text{ AM}=8.3\times10^{-24}Carat \\ \frac{1}{8.3\times10^{-24}}=\frac{8.3\times10^{-24}}{8.3\times10^{-24}}\text{Carat} \\ \frac{1}{8.3\times\frac{1}{10^{24}}}=1\text{Carat} \end{gathered}[/tex]We shall refine the left side of the equation and we'll have;
[tex]\begin{gathered} \frac{1}{1}\div\frac{8.3}{10^{24}} \\ =\frac{1}{1}\times\frac{10^{24}}{8.3} \\ =1.204819277\times10^{23} \end{gathered}[/tex]Therefore;
[tex]1\text{Carat}=1.2\times10^{23}[/tex]For the second part:
[tex]1AM=1.66\times10^{-21}mg[/tex]We now divide the quantity in atomic mass by the quantity in milligrams and that gives us;
[tex]\begin{gathered} \frac{1}{1.66\times10^{-21}}=\frac{1.66\times10^{-21}}{1.66\times10^{-21}}mg \\ \frac{1}{1.66\times\frac{1}{10^{21}}}=1mg \end{gathered}[/tex]We will refine the left side and we'll have;
[tex]\begin{gathered} \frac{1}{1}\div\frac{1.66}{10^{21}}=1mg \\ 1\times\frac{10^{21}}{1.66}=1mg \\ 6.0240963855\times10^{20}=1mg \end{gathered}[/tex]ANSWER:
1 Carat equals;
[tex]\begin{gathered} 1.2\times10^{23} \\ \text{Option B} \end{gathered}[/tex]1 milligram equals;
[tex]\begin{gathered} 6.02\times10^{20} \\ \text{Option D} \end{gathered}[/tex]Find the average rate of change of g(x) = 7/x over the interval [3,6]
Given:
The function is,
[tex]g(x)=\frac{7}{x}\text{ over interval \lbrack{}3,6\rbrack}[/tex]The average rate of change is calcuated as,
[tex]\begin{gathered} \frac{f(b)-f(a)}{b-a} \\ \lbrack a,b\rbrack=\lbrack3,6\rbrack \\ \frac{f(6)-f(3)}{6-3}=\frac{\frac{7}{6}-\frac{7}{3}}{3}=\frac{\frac{21-42}{18}}{3}=\frac{-21}{18\times3}=-\frac{7}{18} \end{gathered}[/tex]Answer: avarage rate of change is -7/18.
A.) (5,8)B.) (-7,-7)C.) (6,-3)D.) (4,4)Are the available options, thank you for the help.
We can find the point that is a solution to this system of linear equations by looking at the shaded region of the graph and determining which of the following points is included inside the shaded region.
By looking at the graph, we can determine that (5,8), (-7,-7), and (6,-3) are not part of the shaded region.
The point (4,4) is the only point that is included in the shaded region, inclusive on the red line.
Therefore, D) (4,4) is a solution to this system of linear inequalities.
Would the answer to this be 5 or did I do something wrong
Simplify the expression
[tex]\begin{gathered} \frac{(3^2+4^2)}{(2+3)}=\frac{9+16}{5} \\ =\frac{25}{5} \\ =5 \end{gathered}[/tex]Answer is 5.
i dont understand conjecture 7:30 7:55 8:20 8:45
conjecture here is nothing but the addition of 25 minutes in the previous time to get next time slot.
[tex]\begin{gathered} \text{ So,} \\ 7\colon30+\text{ 25 =7:55} \\ 7\colon55\text{ +25 = 8:00+20 =8:20} \\ 8\colon20+\text{ 25 = 8:45} \\ 8\colon45+\text{ 25 =9:00+10 = 9:10} \\ 9\colon10\text{ + 25 = 9:35} \end{gathered}[/tex]Write the following sentence as an inequality:"Twice a number is at least 15"
In order to write this sentence as an inequality, let's write each part as a mathematical expression:
"A number" = x
"Twice a number" = 2x
"At least 15" = ≥15
So the final inequality is:
[tex]2x\geq15[/tex]What is the surface area of each figure? Round to the nearest tenth if necessary. A 12 ft 6 ft 8 ft O a. 336 ft2 b. 288 ft? c. 144 ft Od. 134 ft?
Answer: Surface area = 336 ft^2
Explanation:
To calculate the surface area of the given figure, we would find the area of each face of the figure. Looking at the figure, it has
2 triangular faces
one small rectangular face
2 rectangular faces at the top and base
Considering the triangular faces, the formula for calculating the area of a triangle is expressed as
Area = 1/2 x base x height
base = 8
height = 6
Area of 2 triangles = 2 x 1/2 x 8 x 6 = 48
Considering the small rectangle,
length = 12
width = 6
Area of small rectangle = 12 x 6 = 72
Considering the rectangle at the top,
length = 12
width = 10
Area of rectangle at the top = x 12 x 10 = 120
Considering the bigger rectangle at the base,
length = 12
width = 8
Area of rectangle at the top = 12 x 8 = 96
Thus,
Surface area = 48 + 72 + 120 + 96 =
Surface area = 336 ft^2
Each student in the Junior class was asked if they had to complete chores at home and if theyhad a curfew. The table represents the data.Yes choresNo choresTotalYes curfew513081Let:Event A Having a curfewWEvent B Having choresWNo curfew Total24 7512 4236117Which statement accurately describes this data?
SOLUTION
From the table, we can see that the event having a curfew and having chores are dependent events. So there are not independent events.
Also the probability of A given B is
[tex]P(A|B)=\frac{51}{75}[/tex]and Probability of A
[tex]P(A)=\frac{81}{117}[/tex]Hence the second op
fill in the blank. The_ of an event happening is expressed as a fraction between 0 and 1.
the probability of an event happening is expressed as a fraction
mrs.williams use the following recipe for rice crispy. she decides to make 2/3 of the recipe. how much melted butter will she need
Answer
(4/3) cups of melted butter = 1⅓ cups of melted butter.
16 oz of marshmallows.
Explanation
2 cups of melted butter
24 oz marshmallows
13 cups rice crispy cereal
1 full recipe uses 2 cups of melted butter
(2/3) of the recipe will use (2/3) × 2 cups of melted butter = (4/3) cups of melted butter = 1⅓ cups of melted butter.
1 full recipe uses 24 oz of marshmallows
(2/3) of the recipe will use (2/3) × 24 oz of marshmallows = 16 oz of marshmallows.
Hope this Helps!!!
Using the rules of reflections, find the new vertices of the figure reflected across the given line.1. A(2,3), B(-1,5), C(4,-1); y = x2. U(-8,2), V(-3,-1), W(3,3); y-axis 3. E(-3,-2), F(6,-4), G(-2,1); x-axis
1. A(2,3), B(-1,5), C(4,-1); y = x
2. U(-8,2), V(-3,-1), W(3,3); y-axis
3. E(-3,-2), F(6,-4), G(-2,1); x-axis
Part 1
we know that
When you reflect a point across the line y = x, the x-coordinate and y-coordinate change places.
so
the rule is
(x,y) -------> (y,x)
therefore
A(2,3) ------> A'(3,2)
B(-1,5) ------> B'(5,-1)
C(4,-1) ------> C'(-1,4)
Part 2
The rule of the reflection across the y-axis is
(x,y) ----->(-x,y)
therefore
U(-8,2) ------> U'(8,2)
V(-3,-1) -----> V'(3,-1)
W(3,3) -----> W'(-3,3)
Part 3
The rule of the reflection across the x-axis is
(x,y) ------> (x,-y)
therefore
E(-3,-2) -----> E'(-3,2)
F(6,-4) ------> F'(6,4)
G(-2,1) -----> G'(-2,-1)
A cubical water tank has a height of 9.6 inches. How much water canthe tank hold?Round your answer to tenth place.
The volume of a cube is given by:
[tex]V=l^3[/tex]where l is the length of its edges. In this case the length is 9.6 inches, plugging these values we have:
[tex]\begin{gathered} V=(9.6)^3 \\ V=884.736 \end{gathered}[/tex]Therefore, the tank can hold 884.7 cubic inches of water.
Hey I was doing a math assignment and i need your help
Answer
[tex]6\frac{1}{3}\text{ }[/tex]Explanation
The given scale factor of Abraham Lincoln is:
[tex]1\text{ inch }=\frac{4}{9}\text{ f}eet[/tex]The height of Lincoln on the poster is:
[tex]14\frac{1}{4}\text{ inches}[/tex]What to find:
Lincoln's actual height in feet.
Let Lincoln's actual height in feet be represented as x:
[tex]\begin{gathered} \text{Since 1 inch }=\frac{4}{9}\text{ f}eet \\ \text{then }14\frac{1}{4}\text{ inches }=x\text{ f}eet \end{gathered}[/tex]To get the value of x, cross multiply:
[tex]\begin{gathered} x\times1=14\frac{1}{4}\times\frac{4}{9} \\ x=\frac{57}{4}\times\frac{4}{9} \\ x=\frac{57}{9} \\ x=\frac{19}{3} \\ x=6\frac{1}{3}\text{ fe}et \end{gathered}[/tex]Rogrigo practiced the guitar 14 1/3 hours over the past 3 weeks. He practiced for 5 1/4 hours during the first week and 3 2/3 hours during the second week. How much time did Rodrigo spend practicing during the third week?
Total hours spent on practicing the guitar for 3 weeks = 14 1/3 hours
Hours spent in the first week = 5 1/4 hours
Hours spent in the second week = 3 2/3 hours
Let the number of hours spent during the third week be = y
5 1/4 + 3 2/3 + y = 14 1/3
[tex]\begin{gathered} 5\frac{1}{4}\text{ + 3}\frac{2}{3}\text{ + y = 14}\frac{1}{3} \\ \\ \frac{21}{4}+\text{ }\frac{11}{3}\text{ + y =}\frac{43}{3} \\ \\ \frac{63\text{ + 44}}{12}+\text{ y =}\frac{43}{3} \\ \\ \frac{107}{12}\text{ + y =}\frac{43}{3} \\ \\ y\text{ = }\frac{43}{3}\text{ -}\frac{107}{12} \\ \\ y\text{ = }\frac{172\text{ -107}}{12} \\ \\ y\text{ =}\frac{65}{12} \\ \\ y\text{ =5}\frac{5}{12} \end{gathered}[/tex]
What is the length of EC round to the nearest tenth of a foot
we must use trigonometric ratios, which applies for right triangles
on this case we will use
[tex]\tan (\alpha)=\frac{\text{opposite}}{\text{adjacent}}[/tex]where alpha is the angle, opposite is the opposite side of the angle and adjacent the andjacent side of the angle different to hypotenuse
so replacing
[tex]\tan (44)=\frac{x}{2.8}[/tex]and solving for x
[tex]\begin{gathered} x=2.8\tan (44) \\ x=2.7039\approx2.7 \end{gathered}[/tex]the measure of x-side is 2.7 ft
Learning Diagnostic Analytics Recommendations Skill plans Math Level K HH.5 Trigonometric ratios: find an angle measure RSS You have prize Find mZH. I 8 G 4 H Write your answer as an integer or as a decimal rounded to the nearest tenth. mZHE
The Solution:
Given:
We are required to find the measure of angle H.
By applying the trigonometric ratio, we have that:
[tex]\begin{gathered} \tan\angle H=\frac{opposite}{adjacent} \\ \\ \tan\angle H=\frac{8}{4} \end{gathered}[/tex][tex]\begin{gathered} \tan\angle H=2 \\ \text{ Take the arctan of 2} \end{gathered}[/tex][tex]\angle H=\tan^{-1}(2)=63.4349\approx63.4^o[/tex]Therefore, the correct answer is 63.4 degrees.
The sum of a number and it’s square is 30 Find the numbers
The sum of a number and it’s square is 30
[tex]x+x^2=30[/tex]Steps to find x.
1. subtract 30 from both sides and simplify
[tex]\begin{gathered} x+x^2-30=30-30 \\ x^2+x-30=0 \end{gathered}[/tex]2. use the quadratic formula
[tex]x=\frac{-b\pm\sqrt[]{b^2-4\cdot\: a\cdot c}}{2\cdot\: a}[/tex][tex]x_{1,\: 2}=\frac{-1\pm\sqrt{1^2-4\cdot\:1\cdot\left(-30\right)}}{2\cdot\:1}[/tex]3. Solve for the positive value
[tex]x_1=\frac{-1+11}{2\cdot\:1}=5[/tex]4. so the answer is number 5
Write a function rule for the area of rectangle whose length is 2ft more than its width. What is the area of the rectangle when its width is 11ft
According to the problem, the length is 2 feet more than its width, which is expressed as the following function
[tex]l=w+2[/tex]If the width is x, then the length is x+2.
We can express the following function for the area
[tex]A(x)=x(x+2)[/tex]Let's replace the width x = 11 ft.
[tex]A(11)=11\cdot(11+2)=11(13)=143[/tex]Hence, the area is 143 square feet. The function for the area is[tex]A(x)=x(x+2)[/tex]Where x represents the width.
What are the coordinates for the center of the circle and the length of the radius
Given:
[tex]x^2+y^2+14x+2y+14=0[/tex]To find the center of the circle and the radius of the circle:
The given equation is of the form,
[tex]x^2+y^2+2gx+2fy+c=0[/tex]Comparing we get,
[tex]\begin{gathered} g=7 \\ f=1 \\ c=14 \end{gathered}[/tex]The center is,
[tex](-g,-f)=(-7,-1)[/tex]The radius is,
[tex]\begin{gathered} r=\sqrt[]{g^2+f^2-c} \\ =\sqrt[]{(7)^2+(1)^2-14} \\ =\sqrt[]{49+1-14} \\ =\sqrt[]{36} \\ =6\text{ units} \end{gathered}[/tex]Therefore, the center of the circle is (-7,-1) and the radius of the circle is 6 units.
Rome question and type your response in the box provided. Your response will be saved automatically,The wheels on a car has a diameter of 28 inches. Find the area and Circumfrence of the car tire. (YOU MUST FIND BOTH TO GET FULL CREIDT).B IV 112021 Illuminate Education Inc.
Circumference: 28π inches and Area: 196π in²
1) Considering that the diameter of each wheel is 28", then we can state that
D=2R ⇒ R = D/2 , R = 14
2) So to find out the circumference of the tire, we'll need to use the Circumference formula:
[tex]\begin{gathered} C=2\pi\cdot r \\ C=2\pi\cdot14 \\ C=28\pi \end{gathered}[/tex]And now let's find out the area of the tire, by making use of another formula:
Plugging into that the radius length:
[tex]\begin{gathered} A=\pi\cdot r^2 \\ A=(14)^2\pi \\ A=196\pi \end{gathered}[/tex]3) Hence, the answers are:
Circumference: 28π inches and Area: 196π in²
Which set of numbers can represent the lengths of the sides of a triangle ?
It's important to know that the sum of two sides must be greater than the length of the third side, this rule defines triangles. So, we have to make sure that the correct answers meets the rule. For example, the first set of numbers represent a triangle because 3+5 > 7, 3+7 > 5, and 5+7 >3.
Therefore, the correct answer is A.Solve the following system of equations without graphing: 72 + 3y = 53 - 4x + 10y = 29 ) Write your solution as an ordered pair ( (Remember: you want to eliminate one of the variables)
1 72x + 3y= 53
2 -4x + 10y= 29
We are going to multiply our second equation by 18, in order to change the coefficient of x .
2 18*-4x + 18*10y= 29*18
2 -72x + 180y = 522
Then we add the last equation to the first equation
1 72x + 3y= 53
+
2 -72x + 180y = 522
----------------------------------
183 y = 575
y= 575/183
Solving for y we get y= 3.14
Then we replace y in the equation 2 to get x. but first we need to isolate x
2 -4x + 10y= 29
2 -4x = 29 - 10y
2 x = (29 - 10y)/-4
2 x = 0.605
The final answer must be ( 0.605 , 3.14 )
The two legs of a right triangle are the same length. The hypotenuse is 3 meters long. Find the length of the legs, Express your answer insimplified radical form, or as a decimal rounded to four places.
Solution
To solve this question, we will apply Pythagoras theorem, which states that:
[tex]Hypotenuse^2=Opposite^2+Adjacent^2[/tex]- We are given that:
[tex]\begin{gathered} Hypotenuse=3 \\ Opposite=Adjacent=x \end{gathered}[/tex]- Thus, we can say:
[tex]\begin{gathered} 3^2=x^2+x^2 \\ 2x^2=9 \\ \text{ Divide both sides by 2} \\ x^2=\frac{9}{2} \\ \\ Take\text{ the square root of both sides} \\ x=\sqrt{\frac{9}{2}}=\frac{3}{\sqrt{2}} \\ \\ x=\frac{3}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}} \\ \\ x=\frac{3}{2}\sqrt{2} \end{gathered}[/tex]Final Answer
The length of the legs are:
[tex]\frac{3}{2}\sqrt{2}[/tex]