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Alisiya [41]
3 years ago
11

Long division of 7/20​

Mathematics
1 answer:
Nadya [2.5K]3 years ago
8 0
0 .35 Arr u joking u could have just googled it
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Liter of water i drink 1/4 before lunch and 2/3 of the rest after lunch how much do I bring in-between
tresset_1 [31]

Answer:

We have (2/3)x(1-1/4) - 1/4 = (2/3)x(3/4) -1/4 = 6/12 = 1/4 = 1/2 -1/4 = 2/4 - 1/4 = 1/4 do I bring in between;

Step-by-step explanation:

8 0
3 years ago
What is the common point between lines y = ½ x + 4 and y = x – 4?
nirvana33 [79]
1/2X+4=X-4
1/2X+8=X
8=1/2X
16=X

(Pick one of the equations )
y=1/2X+4
y=(1/2)(16)+4
y=12

point: (16,12)
7 0
3 years ago
BRAINLIEST ANSWER UP FOR GRABS!!
Svetach [21]
X = sqrt(5^2 - 3^2) = 4
y = sqrt(7^2 - 3^2) = sqrt 40

Horizontal distance = x + y + 3 = 7 + sqrt40
8 0
3 years ago
Read 2 more answers
Find the mean of:<br> 16 32 10 28 8 19 32 26<br> 33 39 98 399
iren2701 [21]

Answer:

91.125

Step-by-step explanation:

Total number/ Total number of quantity

16+32+10+28+8+19+32+26+33+39+98+399

729

Divode by number of quantity (8)

729/8

=91.125

Hope this helps, dont hesitate to Ask in comment below. Mark me as brainliest also appriciated!!

5 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
Read 2 more answers
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