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skelet666 [1.2K]

3 years ago

15

I have a square piece of paper. Square has all equal sides (each side was 10 cm and there were 4 sides) I fold square A in half

to make rectangle B. I fold rectangle B in half to make square C. a) what is the perimeter of rectangle B? b) what is the perimeter of square C?

1 answer:

Fantom [35]3 years ago

4 0

Answer:

two sides of square c times those to each other

Step-by-step explanation:

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antoniya [11.8K]

Answer:

1. the 3rd one

2. -9

3. 15

4. -8

5 0

3 years ago

Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

Solve

Let

$u = \frac{x^2}{4}$

Differentiate

$du = 2 * \frac{x^{2-1}}{4}\ dx$

$du = 2 * \frac{x}{4}\ dx$

$du = \frac{x}{2}\ dx$

Make dx the subject

$dx = \frac{2}{x}\ du$

The given integral becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

Recall that: $u = \frac{x^2}{4}$

Make $x^2$ the subject

$x^2= 4u$

Square both sides

$x^4= (4u)^2$

$x^4= 16u^2$

Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du$

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

4 0

3 years ago

Let f(x) = 3x3 − 7x2 − 9x + 21 and g(x) = 3x − 7. Find f of x over g of x.

Ronch [10]

Answer:

x^2-3

Step-by-step explanation:

look at the picture

8 0

3 years ago

Read 2 more answers

you're 495 miles from home and you are driving toward home at an average of 45mph. How long will it take to drive home if you ma

LenaWriter [7]

11 hours because I did the math

7 0

3 years ago

Real World Percentages: Roman went shopping for school

Crank

Roman will have to pay $89.67 for his school supplies.
Since you needed help, I’ll show you my work so you know for next time:

85.40 (.05) = 4.27
85.40 + 4.27
= 89.67

I hope this helps :)

8 0

3 years ago