A picture frame (see figure) has a total perimeter of 3 feet. The width of the frame is 0.64 times its length. Find the dimensions of the frame. (Round your answers to two decimal places.)

Answer:

100

Step-by-step explanation:

I believe it is this answer bur don't hold me to. If it is wrong I'm sorry in advance

Answer:

The two angles are 31° and 149°

Step-by-step explanation:

Let the two supplementary angles be , x and y.

         Then, x + y = 180° ----- ( 1 )

Given :  one is 25° more than 4 times other

           Let y = 25° + 4x ------ ( 2 )

Substitute ( 2 ) in  ( 1 ) :

     x + y = 180°

     x + 25° +4x = 180°

     x + 4x = 180° - 25°

           5x = 155°

             x = 31°

Substitute x in ( 2 ) :

            y = 25° + 4x

               = 25 + 4 ( 31 )

               = 25 + 124

               = 149°

Answer:

No, 6 hours would cost $335, 35 over her limit of $300.

Step-by-step explanation:

she has to pay $200 plus 22.50 for every hour she rents. So, if she rents for 6 hours, she has to pay $200 + 6(22.50) or $200 + 135=335. She only has $300 so she can’t afford the event.

Answer:

A

Step-by-step explanation:

Answer: c

Step-by-step explanation: edge 2021

x = 65

Step-by-step explanation:

cos theta = 25/x

cos theta = x/169

25/x = x/169

x² = 169 x 25

x = 65

The missing length x = 65, using the Pythagoras Theorem.

According to the Pythagoras Theorem, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs.

In the question, we are asked to find the value of x.

In the right triangle ABC, by Pythagoras' Theorem,

AC² + BC² = AB²,

or, x² + BC² = (144 + 25)²,

or, BC² = 169² - x² ... (i).

In the right triangle ACD, by Pythagoras Theorem,

AD² + DC² = AC²,

or, 25² + DC² = x²,

or, DC² = x² - 25² ... (ii).

In the right triangle BCD, by Pythagoras Theorem,

BD² + DC² = BC²,

or, 144² + x² - 25² = 169² - x² {Substituting BC² = 169² - x² from (i) and DC² = x² - 25² from (ii)},

or, x² + x² = 169² + 25² - 144² {Rearranging},

or, 2x² = 28561 + 625 - 20736,

or, 2x² = 8450,

or, x² = 4225,

or, x = √4225 = 65.

Thus, the missing length x = 65, using the Pythagoras Theorem.

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Answer:

m=2, n=-1 or (2,-1)

Step-by-step explanation:

Hi there!

We're given this system of equations:

3n+5m=7

m-4n=6

we want to solve by substitution; in one equation, we'll set one variable to equal an expression containing the other variable, substitute that expression as the variable into the other equation, solve for the variable substituted, and use the value of that variable to find the other variable

First, let's get the expression we'll substitute into the other

let's use m-4n=6 as it has 1m (m by itself)

add 4n to both sides

m=4n+6

now substitute 4n+6 as m into 3n+5m=7 to solve for n

3n+5(4n+6)=7

do the distributive property

3n+20n+30=7

combine like terms

23n+30=7

subtract 30 from both sides

23n=-23

divide both sides by 23

n=-1

we found the value of n

now let's substitute -1 as n in m=4n+6 to find the value of m

m=4(-1)+6

multiply

m=-4+6

add

m=2

so the answer is m=2, n=-1. As a point, it's (2,-1)

Hope this helps! :)

Answer:

787.50

Step-by-step explanation:

interest=principle*rate*time/100

interest=3500*7.5*3/100

interest=35*7.5*3

interest=787.50

Answer:

B. $580

Step-by-step explanation:

Value of the retirement fund after x years:

The value of the retirement fund after x years is given by the following function:

[tex]y(x) = 307.6(1.11)^x[/tex]

Which is best estimate for the value of the retirement fund after 6 years, if no additional deposits or withdrawals are made?

This is y(6). So

[tex]y(6) = 307.6(1.11)^6 = 575.34[/tex]

Close to $580, and thus, the correct answer is given by option B.

Answer:

B. $580 is best estimate for the value of the retirement fund after 6 years, if no additional deposits or withdrawals are made.

Thus it is little close to answer because actual answer is $575.34.

Answer:

r = 22.5

Step-by-step explanation:

[tex]V=\frac{4}{3} \pi r^{3}[/tex]

[tex]11411=\frac{4}{3} \pi r^{3}[/tex]

[tex]r^{3}=3(\frac{11411}{3})[/tex]

[tex]r^{3}=11411[/tex]

∛r³ = ∛11411

r = 22.5134076467

22.5134076467 rounded to the nearest tenth is 22.5

Answer:

its 14

Step-by-step explanation:

Answer:

2.3

Step-by-step explanation:

A number when squarerooted squared will be come that number.

In formula terms, it is something like that:

[tex]\sqrt{x} X \sqrt{x} =x[/tex]

So…

[tex]\sqrt{2.3} X \sqrt{2.3} = 2.3[/tex]

Answer:

6

Step-by-step explanation:

Answer:

Q1.

a. y = 3x + 0

b. y = 3x + 25

c. Plain is proportional because the y-intercept is zero.

d. It shows the cost of each coaster.

Q2

a. y = 4x + 0

b. It is proportional since the y-intercept is zero.

c. No the graph will never intersect because the equations in combination will result in parallel lines since there is a slight variation in the cost of EZ Coaters

Answer:

-3.5, -3, -1, 3, 4.5

Step-by-step explanation:

Answer:

-3.5, -3, -1, 3, 4.5 because the higher the negative is, the lower the number.

Answer:

[tex]Cov(X,Y) = -\frac{ 25}{9}[/tex]

Step-by-step explanation:

Given

[tex]Interval =[0,10][/tex]

[tex]X + Y < 10[/tex]

Required

[tex]Cov(X,Y)[/tex]

First, we calculate the joint distribution of X and Y

Plot [tex]X + Y < 10[/tex]

So, the joint pdf is:

[tex]f(X,Y) = \frac{1}{Area}[/tex] --- i.e. the area of the shaded region

The shaded area is a triangle that has: height = 10; width = 10

So, we have:

[tex]f(X,Y) = \frac{1}{0.5 * 10 * 10}[/tex]

[tex]f(X,Y) = \frac{1}{50}[/tex]

[tex]Cov(X,Y)[/tex] is calculated as:

[tex]Cov(X,Y) = E(XY) - E(X) \cdot E(Y)[/tex]

Calculate E(XY)

[tex]E(XY) =\int\limits^X_0 {\int\limits^Y_0 {\frac{XY}{50}} \, dY} \, dX[/tex]

[tex]X + Y < 10[/tex]

Make Y the subject

[tex]Y < 10 - X[/tex]

So, we have:

[tex]E(XY) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{XY}{50}} \, dY} \, dX[/tex]

Rewrite as:

[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {XY}} \, dY} \, dX[/tex]

Integrate Y

[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{XY^2}{2}}} }|\limits^{10 - X}_0 \, dX[/tex]

Expand

[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{X(10 - X)^2}{2} - \frac{X(0)^2}{2}}} }\ dX[/tex]

[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{X(10 - X)^2}{2}}} }\ dX[/tex]

Rewrite as:

[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 X(10 - X)^2\ dX[/tex]

Expand

[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 X*(100 - 20X + X^2)\ dX[/tex]

[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 100X - 20X^2 + X^3\ dX[/tex]

Integrate

[tex]E(XY) =\frac{1}{100} [\frac{100X^2}{2} - \frac{20X^3}{3} + \frac{X^4}{4}]|\limits^{10}_0[/tex]

Expand

[tex]E(XY) =\frac{1}{100} ([\frac{100*10^2}{2} - \frac{20*10^3}{3} + \frac{10^4}{4}] - [\frac{100*0^2}{2} - \frac{20*0^3}{3} + \frac{0^4}{4}])[/tex]

[tex]E(XY) =\frac{1}{100} ([\frac{10000}{2} - \frac{20000}{3} + \frac{10000}{4}] - 0)[/tex]

[tex]E(XY) =\frac{1}{100} ([5000 - \frac{20000}{3} + 2500])[/tex]

[tex]E(XY) =50 - \frac{200}{3} + 25[/tex]

Take LCM

[tex]E(XY) = \frac{150-200+75}{3}[/tex]

[tex]E(XY) = \frac{25}{3}[/tex]

Calculate E(X)

[tex]E(X) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{X}{50}}} \, dY} \, dX[/tex]

Rewrite as:

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {X}} \, dY} \, dX[/tex]

Integrate Y

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 { (X*Y)|\limits^{10 - X}_0 \, dX[/tex]

Expand

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 ( [X*(10 - X)] - [X * 0])\ dX[/tex]

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 ( [X*(10 - X)]\ dX[/tex]

[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 10X - X^2\ dX[/tex]

Integrate

[tex]E(X) =\frac{1}{50}(5X^2 - \frac{1}{3}X^3)|\limits^{10}_0[/tex]

Expand

[tex]E(X) =\frac{1}{50}[(5*10^2 - \frac{1}{3}*10^3)-(5*0^2 - \frac{1}{3}*0^3)][/tex]

[tex]E(X) =\frac{1}{50}[5*100 - \frac{1}{3}*10^3][/tex]

[tex]E(X) =\frac{1}{50}[500 - \frac{1000}{3}][/tex]

[tex]E(X) = 10- \frac{20}{3}[/tex]

Take LCM

[tex]E(X) = \frac{30-20}{3}[/tex]

[tex]E(X) = \frac{10}{3}[/tex]

Calculate E(Y)

[tex]E(Y) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{Y}{50}}} \, dY} \, dX[/tex]

Rewrite as:

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {Y}} \, dY} \, dX[/tex]

Integrate Y

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 { (\frac{Y^2}{2})|\limits^{10 - X}_0 \, dX[/tex]

Expand

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 ( [\frac{(10 - X)^2}{2}] - [\frac{(0)^2}{2}])\ dX[/tex]

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 ( [\frac{(10 - X)^2}{2}] )\ dX[/tex]

[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 [\frac{100 - 20X + X^2}{2}] \ dX[/tex]

Rewrite as:

[tex]E(Y) =\frac{1}{100}\int\limits^{10}_0 [100 - 20X + X^2] \ dX[/tex]

Integrate

[tex]E(Y) =\frac{1}{100}( [100X - 10X^2 + \frac{1}{3}X^3]|\limits^{10}_0)[/tex]

Expand

[tex]E(Y) =\frac{1}{100}( [100*10 - 10*10^2 + \frac{1}{3}*10^3] -[100*0 - 10*0^2 + \frac{1}{3}*0^3] )[/tex]

[tex]E(Y) =\frac{1}{100}[100*10 - 10*10^2 + \frac{1}{3}*10^3][/tex]

[tex]E(Y) =10 - 10 + \frac{1}{3}*10[/tex]

[tex]E(Y) =\frac{10}{3}[/tex]

Recall that:

[tex]Cov(X,Y) = E(XY) - E(X) \cdot E(Y)[/tex]

[tex]Cov(X,Y) = \frac{25}{3} - \frac{10}{3}*\frac{10}{3}[/tex]

[tex]Cov(X,Y) = \frac{25}{3} - \frac{100}{9}[/tex]

Take LCM

[tex]Cov(X,Y) = \frac{75- 100}{9}[/tex]

[tex]Cov(X,Y) = -\frac{ 25}{9}[/tex]

Answer:

Step-by-step explanation:

Answer:

B) Coefficient.

Step-by-step explanation:

The coefficient is the numerical constant in the given term.

For example, the given term is 4x, in which:

4 is the coefficient.

x is the variable.

5 is the constant.

20 is another constant.

~

Answer:

[tex]x = -6[/tex]

Step-by-step explanation:

[tex]18 = [(-2) + (-1)]x[/tex]

Required

Find x

Remove the inner brackets

[tex]18 = [-2-1]x[/tex]

[tex]18 = [-3]x[/tex]

Remove the square bracket

[tex]18 = -3x[/tex]

Divide both sides by -3

[tex]\frac{18}{-3} = x[/tex]

[tex]x = -6[/tex]

Answer:

I think it’s 2 hope my answer was good have a nice day as well

Step-by-step explanation:

The period of the given function y = 5 sin (πx) + 4 is π.

We have given that,

y = 5 sin (πx) + 4

We have to determine the period

The period of the sine curve is the length of one cycle of the curve. The natural period of the sine curve is 2π.

Therefore the period of the given function is π.

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Answer:

m∠A = 32°

Step-by-step explanation:

8^2 = 12^2 + 15^2 - 2*12*15*cosA

Answer:

B

Step-by-step explanation:

did it on edge

Answer:

no...you can just use a clip on

Answer:

it's absolutely worth it. The pain of getting one isn't that bad, at least for me. It's just a quick pinch. It'll look so good. Just make sure you are also cleaning it.

Step-by-step explanation:

Answer:

Graph 4

Step-by-step explanation:

Answer:

I think the answer is 1^2 - 10

[tex] \orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}} [/tex]

FIND THE NUMBER THAT MAKE EQUIVALENT FRACTION.

[tex] \huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}[/tex]

[tex] \red{solved} \\ \frac{4}{ \cancel{5}} = \frac{x}{ \cancel{40}^{ \: \: \: \tiny{8}} } \\ 8 \times 4 = x \\ 32 = x[/tex]

[tex] \blue{Part - 1} \\ \frac{4}{ \cancel{6}} = \frac{x}{ \cancel{24}^{ \: \: \: \tiny{4}} } \\ 4\times 4 = x \\ 16 = x[/tex]

[tex] \green{Part - 2} \\ \frac{1}{ 2} = \frac{6}{ x } \\ 6 \times 2 = x \\ 12 = x[/tex]

SOLVE LIKE THIS FOR ALL

PART - 3 --> 12

PART - 4--> 40

PART - 5 --> 6

PART - 6--> 4

PART - 7 --> 54

PART - 8 --> 12

PART - 9--> 36

PART - 10 --> 28

PART - 11--> 20

PART - 12 --> 28

PART - 13 --> 16

PART - 14 --> 20

PART - 15 --> 12

PART - 16 --> 6

PART - 17 --> 32

PART - 18--> 30

PART - 19 --> 27

PART - 20 --> 36

Answer:

[tex]F = k\frac{m_1m_2}{r^2}[/tex]

Step-by-step explanation:

Direct proportional:

Multiplication

Inverse proportional:

Division

The gravitational attraction F between two objects of masses m1 and m2 is jointly proportional to the masses

This means that:

[tex]F = k\frac{m_1m_2}{}[/tex]

Inversely proportional to the square of the distance r between the objects.

Denominator is [tex]r^2[/tex]. Thus:

[tex]F = k\frac{m_1m_2}{r^2}[/tex]

Explanation:

The sample space is simply the list of possible outcomes. We list all the hair colors possible in this class.

Answer:

A

Step-by-step explanation:

Answer:

Step-by-step explanation:

y=5px+5x2+p2

y=10p2x2p

2x²×(2x²+3)=2-x²

[tex]x = \frac{1}{2} , - \frac{1}{2} ,i \sqrt{2} , - i \sqrt{2} [/tex]