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3 years ago

8

May runs10/4 miles everyday after school .Beth says that she runs farther then May everyday 5/2 miles after school is her correc

t? Explain your ansewr

1 answer:

Shalnov [3]3 years ago

5 0

They run an equally amount of time because 10/4 = 2 and 1/2 and 5/2 = 2 and 1/2

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The product of two consecutive positive integers is 272. What is the sum of the two integers?

Mekhanik [1.2K]

Answer:

the sum of two integers is 33

explanation:

the numbers are

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

Can someone help me please

miss Akunina [59]

15 is A, 22+X=8
I don't have time to solve the rest RN but will come back later

3 0

3 years ago

Read 2 more answers

Simplify:<br> 4n + 6 - (5n+3)

Rudik [331]

Answer:

-n+3

Step-by-step explanation:

Subtract the 5n and 3

6 0

3 years ago

Help and please explain so I can do it on my own. There are a couple more on this sheet.

Andreas93 [3]

Answer:

(2, -10)

Step-by-step explanation:

so first i find the axis of symmetry using the equation x= -b/2a

in this problem

a=3

b= -12

c= 2

- (-12)/ 2(3) = 2

2 will be the x point for your vertex

you then plug in your x point into the original equation and solve for y

y = 3(2)^2 - 12(2) +2

y = -10

Vertex is then (2, -10)

8 0

3 years ago