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2 years ago

Which of the following tables represent functions? If the relation is not a function, indicate why it is not.

Mathematics

1 answer:

Varvara68 [4.7K]2 years ago

6 0

Step-by-step explanation:

1. the table on the left is a function, with the formula : y = x²

2. the table on the right is also a function, indicated by the values of the input are different each others

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There are 75 students enrolled in a camp. The day before the camp begins, 8% of the students cancel. How many students actually

8090 [49]

Answer:

69 students attend the camp

Step-by-step explanation:

8% of 75 = 6

75 - 6 = 69

6 0

2 years ago

Read 2 more answers

Please, I need help in this ??

nignag [31]

Answer:

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

Step-by-step explanation:

$\int\frac{x^{4}}{x^{4} -1}dx$

Adding and Subtracting 1 to the Numerator

$\int\frac{x^{4} - 1 + 1}{x^{4} -1}dx$

Dividing Numerator seperately by $x^{4} - 1$

$\int 1 + \frac{1}{x^{4}-1 }\, dx$

Here integral of 1 is x +c1 (where c1 is constant of integration

$x + c1 + \int\frac{1}{(x-1)(x+1)(x^{2}+1)}\, dx$ ----------------------------------(1)

We apply method of partial fractions to perform the integral

$\frac{1}{(x-1)(x+1)(x^{2}+1)}$ = $\frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x^{2} + 1}$ ------------------------------------------(2)

$\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)}{(x-1)(x+1)(x^{2} +1)}$

1 = $A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)$ -------------------------(3)

Substitute x= 1 , -1 , i in equation (3)

1 = A(1+1)(1+1)

A = $\frac{1}{4}$

1 = B(-1-1)(1+1)

B = $-\frac{1}{4}$

1 = C(i-1)(i+1)

C = $-\frac{1}{2}$

Substituting A, B, C in equation (2)

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\int\frac{1}{4(x-1)} - \frac{1}{4(x+1)} -\frac{1}{2(x^{2}+1) }$

On integration

Here $\int \frac{1}{x}dx = lnx and \int\frac{1}{x^{2}+1 } dx = arctanx$

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\frac{1}{4} ln(x-1)$ - $\frac{1}{4} ln(x+1)$ - $\frac{1}{2} arctanx$ + c2---------------------------------------(4)

Substitute equation (4) back in equation (1) we get

$x + c1 + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2$

Here c1 + c2 can be added to another and written as c

Therefore,

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

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3 years ago

A street is drawn by dilating segment FG¯ about center A with a scale factor greater than 0 but less than 1. Is this an enlargem

Arte-miy333 [17]

Answer: This is an reduction.

Step-by-step explanation:

A dilation a king of transformation that creates an similar image (about a center of dilation) of the actual figure by changing its size with the use of a scale factor(k).

It either shrinks or stretches the image.

If |k| is greater than 1 then the image is an enlargement .
If |k| is less than 1 then the image is an reduction.
If |k| is equals to 1 then there is no change in size.

Given : A street is drawn by dilating segment $\overline{FG}$ about center A with a scale factor greater than 0 but less than 1.

Then by using (2.) , we can say that this is an reduction.

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In a school of 500 students, 160 have blue eyes, 320 have brown eyes and 20 have green eyes. In a class of 25 students how many

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160/500= 80/250 16/50 8/25

Around 32 percent

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