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Kay [80]
3 years ago
5

Use the net to find the size of the base and the height of each triangular face of this square pyramid.

Mathematics
2 answers:
nlexa [21]3 years ago
7 0
The base would be 16 and the height of each triangle would be 2
antiseptic1488 [7]3 years ago
4 0
The base is 16 and height of each triangle is 2
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Add 4xly, -2xy?, -5xy?, 312y​ <br>step by step explanation plzz
storchak [24]

Answer:

8xy that is the answer...................

8 0
3 years ago
If y=6 when x=10 find x when y=18 equation
egoroff_w [7]
I believe x would equal to 30.

Since 6 * 3 =18, 10 * 3 = 30.
What you do to one side, you must do to the other.

Hope I helped!

6 0
4 years ago
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C(25,10)<br> Find permutation and combination
Verizon [17]

C(25,10)= 3268760

P(25,10)=11861676288000

hope it helps but these are very big numbers :d

6 0
3 years ago
NEED HELP ASAP!!!!!!! WILL MARK CORRECT ANSWER BRAINLIEST!!!!
Ganezh [65]

Answer:

(x + 1)² + (y - 1)² = 1.5625

Step-by-step explanation:

From the picture we see:

The radius is 5 units. Each unit is .25 so the radius is 1.25.

The centre or middle M= ( -1, 1 )

Circle with centre M and radius r :

(x - xM)² + (y - yM)² = r²

xM = -1

yM = 1

r = 1.25

(x - (-1))² + (y - 1)² = (1.25)²

(x + 1)² + (y - 1)² = 1.5625

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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