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AlekseyPX
3 years ago
11

Given the list of x's and y's what is the constant of proportionality. X- 5, 6, 7, 8, 9 Y- 30, 36, 42, 48, 54

Mathematics
1 answer:
Snezhnost [94]3 years ago
7 0

Answer:

Constant of proportionality(X:Y) = 1:6

Step-by-step explanation:

Given:

X- 5, 6, 7, 8, 9

Y- 30, 36, 42, 48, 54

Find:

Constant of proportionality

Computation:

X/Y = 5/30 = 6/36 = 7/42 = 8/48 = 9/54

X/Y = 1/6 = 1/6 = 1/6 = 1/6 = 1/6

So;

Constant of proportionality(X:Y) = 1:6

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- How is the formula for the perimeter of a rectangle an application of the Distributive Property
N76 [4]
Distributive is
a(b+c)=ab+ac
ab+ac=a(b+c)

perimiter of a rectangle=distance around

we have
P=L+W+L+W if we go around
so
P=L+L+W+W
P=2L+2W
undistribute 2
P=2(L+W)
3 0
3 years ago
PLEASE ANSWER
melamori03 [73]

Answer:

A. (2,2), (3,2), (3,3), (3,4), (4,6), (5,4), (5,7), (6,4), (6,6), (7,7)

6 0
4 years ago
Read 2 more answers
how many 5-digit even numbers can be formed with the digits 1,2,3,4,5, and 6 if repetition is allowed?​
Setler79 [48]

Given:

The given digits are 1,2,3,4,5, and 6.

To find:

The number of 5-digit even numbers that can be formed by using the given digits (if repetition is allowed).

Solution:

To form an even number, we need multiples of 2 at ones place.

In the given digits 2,4,6 are even number. So, the possible ways for the ones place is 3.

We have six given digits and repetition is allowed. So, the number of possible ways for each of the remaining four places is 6.

Total number of ways to form a 5 digit even number is:

Total=6\times 6\times 6\times 6\times 3

Total=3888

Therefore, total 3888 five-digit even numbers can be formed by using the given digits if repetition is allowed.

6 0
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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Need help with this one
Simora [160]

Answer:

try c

Step-by-step explanation:

6 0
3 years ago
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