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4 years ago

20，I am not sure how to do it

Mathematics

1 answer:

professor190 [17]4 years ago

5 0

Given:
f(x) = x^4 - x

g(x) = x^4 (even)
h(x) = -x (odd)

f(x) as a sum of an even function and an odd function

f(x) = g(x) + h(x)

s(x) = 1 / (x^4 - x)

s(x) as a sum of an even function and an odd function

s(x) = 1 / [g(x) + h(x)]

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What is the solution to this equation?

tresset_1 [31]

Answer:

x = 9

Step-by-step explanation:

First, we want to isolate variable x on the left side of the equation. We can do this by adding 13 to both sides.

x - 13 + 13 = - 4 + 13

x = 9

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Naily [24]

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BigorU [14]

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the area of a rectangular volleyball court is 1800 square feet the courts length is twice it width write a system of equations t

Marianna [84]

Answer:

The dimensions of the rectangular volleyball court are 60 ft x 30 ft

Step-by-step explanation:

Let

x ----> the length of rectangular volleyball court

y ---> the width of the rectangular volleyball court

we know that

The area of the rectangular volleyball court is equal to

$A=xy$

$A=1,800\ ft^2$

$1,800=xy$ ----> equation A

$x=2y$ -----> equation B

substitute equation B in equation A

$1,800=(2y)y$

$1,800=2y^2$

Solve for y

Simplify

$900=y^2$

take square root both sides

$y=30\ ft$

<em>Find the value of x</em>

$x=2y$

substitute the value of y

$x=2(30)=60\ ft$

therefore

The dimensions of the rectangular volleyball court are 60 ft x 30 ft

7 0

4 years ago

you currently have 24 credit hours and a 2.8 gpa you need a 3.0 gpa to get into the college. if you are taking a 16 credit hours

Juliette [100K]

Answer:

$\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2$

And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:

$\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0$

And we can solve for $\sum_{i=1}^n w_f *X_f$ and solving we got:

$3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f$

And from the previous result we got:

$3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f$

And solving we got:

$\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8$

And then we can find the mean with this formula:

$\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3$

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

Step-by-step explanation:

For this case we know that the currently mean is 2.8 and is given by:

$\bar X = \frac{\sum_{i=1}^n w_i *X_i }{24} = 2.8$

Where $w_i$ represent the number of credits and $X_i$ the grade for each subject. From this case we can find the following sum:

$\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2$

And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:

$\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0$

And we can solve for $\sum_{i=1}^n w_f *X_f$ and solving we got:

$3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f$

And from the previous result we got:

$3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f$

And solving we got:

$\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8$

And then we can find the mean with this formula:

$\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3$

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

6 0

3 years ago