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LUCKY_DIMON [66]
3 years ago
15

A painting is 2 in tall and 3 in wide. If it is enlarged to a width of 15 in then how tall will it be?

Mathematics
2 answers:
vekshin13 years ago
7 0

Answer:

10 in

Step-by-step explanation:

Old proportions:

2 x 3

New proportions

h x 15

Assuming the painting has been increased by a constant scale factor.

Where x is the unknown scale factor, and h is the height after it has been enlarged.

3 * x = 15\\2 * x = h

Rearrange second equation:

3 * x = 15\\x =\frac{h}{2}

Substitute:

3*\frac{h}{2} =15\\\frac{h}{2}=5\\h=10

Papessa [141]3 years ago
7 0

Answer:

hey!

your answer is

10 inches

Step-by-step explanation:

the answer is this because,

15 inches is 3 inches times by 5

so we would have to times 2 inches by 5 to get 10 inches

so the dimensions would be 10 inches in length and 15 inches in width

hope this helps

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Answer:

6

Step-by-step explanation:

You must find the least common multiple of their schedules.

The numbers 3 and 2 are their own prime factors, so the least common multiple of 3 and 2 is 6.

Six weekends must pass until they can both have someone over on the same night.

The number line below shows that the first time their sleepovers coincide is Week 6.

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Alexa is planning to drive across the United States in
ivann1987 [24]

Answer:

a=112

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Step-by-step explanation:

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28x=y

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3 years ago
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How do I solve for AB? (rounded to the nearest tenth)
kolbaska11 [484]

Answer:

The answer would be 8.1

Step-by-step explanation:

For the smaller triangle, you use the pythagorean theorem. a squared + b squared = c squared.

To find one of the legs, you do 5 squared - 3 squared = b squared.

25 - 9 = b squared. (BD)

16 = b squared

4 = b

Now for the bigger right triangle. You still use the same tactic.

7 squared + 4 squared = c squared (which is AB)

49 + 16 = c squared

65 = c squared

That means c would equal: square root of 65, which is 8.0622577483 which rounds to 8.1

7 0
2 years ago
Consider the function f given by f(x)=x*(e^(-x^2)) for all real numbers x.
NISA [10]

Answer:

\frac{\sqrt{\pi}}{4}

Step-by-step explanation:

You are going to integrate the following function:

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furthermore, you know that:

\int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}

lets call to this integral, the integral Io.

for a general form of I you have In:

I_n=\int_0^{\infty}x^ne^{-ax^2}dx

furthermore you use the fact that:

I_n=-\frac{\partial I_{n-2}}{\partial a}

by using this last expression in an iterative way you obtain the following:

\int_0^{\infty}x^{2s}e^{-ax^2}dx=\frac{(2s-1)!!}{2^{s+1}a^s}\sqrt{\frac{\pi}{a}} (2)

with n=2s a even number

for s=1 you have n=2, that is, the function g(x). By using the equation (2) (with a = 1) you finally obtain:

\int_0^{\infty}x^2e^{-x^2}dx=\frac{(2(1)-1)!}{2^{1+1}(1^1)}\sqrt{\pi}=\frac{\sqrt{\pi}}{4}

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