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3 years ago

6

Jonah studied for 3 hours longer than Sierra. he also studied for 3 times as many hours as William. these relationships are show

n on the graphs below. which statement is true if Jonah studied for 6 hours?
A) William studied for 3 hours longer than Sierra
B) Sierra studied 3 hours longer than William
C) William studied for 1 hour longer than Sierra
D) Sierra studied 1 hour longer than William

2 answers:

denpristay [2]3 years ago

8 0

I think it’s A because if you think about it and add it up you should get it

FrozenT [24]3 years ago

6 0

Your answer is A step by step

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Im struggling a bit can someone help me :)

Rasek [7]

Answer:

1

Step-by-step explanation:

6 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

kotykmax [81]

Answer:okay send me a picture of your lesson

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

4) Choose the correct inequality for this situation.

Ksenya-84 [330]

Answer:

B. 21x + 40 ≤ 124

Step-by-step Explanation:

Maximum amount budgeted = $124 (this means they can't spend more than this)

x = number of people

Cost per head = $21

Given that Mr Walter already spent $40, which is part of the money budgeted, the number of people that can go canoeing cam be expressed with the following inequality:

21x + 40 ≤ 124

(note: the amount total to be spent will either be equal to or greater than $124, because it's the maximum amount budgeted for spending).

6 0

3 years ago

Which of the following shows the division problem below in synthetic division form? (6x^3+x^2-3)/(x-7)

PSYCHO15rus [73]

Answer:

Option C.

Step-by-step explanation:

If a polynomial P(x) is divided by (x-c), then we write leading coefficients inside the synthetic division line and c outside the synthetic division line.

The given division problem is

$\dfrac{6x^3+x^2-3}{x-7}$

Here,

$x-c=x-7\Rightarrow c=7$

The numerator polynomial can be written as

$P(x)=6x^3+x^2+0x-3$

So, the coefficients of numerator are 6, 1, 0, -3.

The synthetic division form for given division problem is

7 | 6 1 0 -3

Therefore, the correct option is C.

5 0

3 years ago