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Elena L [17]
2 years ago
5

The scale in the drawing is 2 cm : 5 m. Select from the drop down the correct numbers into the boxes to give the length, width,

and area of the actual room.
Mathematics
1 answer:
Gelneren [198K]2 years ago
7 0

The length, width, and area will be 25m, 15m, and 375m².

<h3>How to calculate the area?</h3>

From the information given, the scale in the drawing is 2cm : 5m and the length and width are given as 10cm and 6cm respectively.

Length = 10cm = 10 × 2.5 = 25m

Width = 6cm = 6 × 2.5m = 15m

Area = Length × Width

Area = 25 × 15

Area = 375m²

Therefore, the area is 375m².

Learn more about area on:

brainly.com/question/25292087

#SPJ1

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3. No it is not on the graph.

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3 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:

g(x)=x*f(x)=x*xe^{-x^2}=x^2e^{-x^2}  (1)

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}

5 0
3 years ago
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irakobra [83]
<span>√9/49

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5 0
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atroni [7]
X+8y=18
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Multiply the first equation by 5
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