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4 years ago

The perimeter of a rectangular parking lot is 292 m.

Mathematics

2 answers:

Vikentia [17]4 years ago

8 0

Answer:

68m

Step-by-step explanation:

The perimeter of a rectangle with length and width is given by the following.

P=21+2w

We need to find w when P=292 m and l+78m .

Using these values in the equation above, we'll solve for w.

P=21+2w

292=2(78)+2w

292=156+2w

136=2w

68=w

Subtracting 156 from both sides

Dividing both sides by 2

So the width of the parking lot is 68m

Reika [66]4 years ago

4 0

Answer:

the answer would be 211

Step-by-step explanation:

you simply add the length and width together to find perimeter. since you have the total perimeter and the length you would instead subtract the two.

292-81= 211

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Use the following graph of the function f(x) = x3 − 3x2 − x + 3 to answer this question:

WITCHER [35]

Answer:

the average rate of change is equal to $0$

Step-by-step explanation:

we have

$f(x)=x^{3}-3x^{2}-x+3$ -----> a cubic function

For $x=-1$

The value of f)x) is equal to

$f(-1)=0$ -----> see the graph

For $x=3$

The value of f(x) is equal to

$f(3)=0$ -----> see the graph

we know that

the average rate of change using the graph is equal to

$\frac{f(b)-f(a)}{b-a}$

In this problem we have

$f(a)=f(-1)=0$

$f(b)=f(3)=0$

$a=-1$

$b=3$

Substitute

$\frac{0-0}{3+1}=0$

8 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=cos%20%5B%20arctan%28%5Cfrac%7B12%7D%7B5%7D%29%20-%20arcsin%20%28%5Cfrac%7B-3%7D%7B5%7D%29%5D"

qwelly [4]

Answer: -16/65

Step-by-step explanation:

Drawing the right triangle (as attached) gives us that $\arctan \left(\frac{12}{5} \right)=\arcsin \left(\frac{12}{13} \right)$

Also, $-\arcsin \left(-\frac{3}{5} \right)=\arcsin \left(\frac{3}{5} \right)$

This means our original expression is equal to:

$\cos \left[\arcsin \left(\frac{12}{13} \right)+\arcsin \left(\frac{3}{5} \right) \right]$

Using the cosine addition formula, which states $\cos(a+b)=\cos a \cos b-\sin a \sin b$ , we get this itself is equal to:

$\cos \left(\arcsin \left(\frac{12}{13} \right) \right)\cos \left(\arcsin \left(\frac{3}{5} \right)\right)-\sin \left(\arcsin \left(\frac{12}{13} \right) \right)\sin \left(\arcsin \left(\frac{3}{5} \right)\right)$

Since $\sin^{2} \theta+\cos^{2} \theta=1$ , we know that:

$\sin^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)+\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\\frac{144}{169} +\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{25}{169}\\\\cos \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{5}{13}$

Similarly, cos(arcsin(3/5))=4/5.

This means the given expression is equal to:

$\left(\frac{5}{13} \right) \left(\frac{4}{5} \right)-\left(\frac{12}{13} \right) \left(\frac{3}{5} \right)\\\\\frac{20}{65}-\frac{36}{65}=\boxed{-\frac{16}{65}}$

3 0

2 years ago

Find the surface area of the prism.

morpeh [17]

B I think I’m not sure because you did not include the type of prism... rectangular triangular? Also if your talking about lateral surface area! so 400mm squared b I think

6 0

3 years ago

An urn contains 2n balls, of which 2 are numbered 1, 2 are numbered 2, ... , and 2 are numbered n. balls are successively withdr

madam [21]

She had a 88% chance because she used enough urn

3 0

3 years ago

Following triangles are similar

oksian1 [2.3K]

Answer:

40m

Step-by-step explanation:

0.2/1=8/x

cross multiply

8*1=0.2x

8=0.2x

8/0.2=x

x=40

3 0

3 years ago