How many of the students correctly applied only the associative property to rewrite the expression?

HELP PLEASEEEE I don’t get it

jek_recluse [69]

10 is answer choice B

11 is answer choice A

7 0

3 years ago

Read 2 more answers

Select all the correct answers. Which two inequalities can be used to find the solution to this absolute value inequality? 3|x +

ziro4ka [17]

The absolute value inequality can be decomposed into two simpler ones.

x < 0

x > -8

<h3></h3><h3>Which two inequalities can be used?</h3>

Here we start with the inequality:

3|x + 4| - 5 < 7

First we need to isolate the absolute value part:

3|x + 4| < 7 + 5

|x + 4| < (7 + 5)/3

|x + 4| < 12/3

|x + 4| < 4

The absolute value inequality can now be decomposed into two simpler ones:

x + 4 < 4

x + 4 > - 4

Solving both of these we get:

x < 4 - 4

x > -4 - 4

x < 0

x > -8

These are the two inequalities.

Learn more about inequalities:

brainly.com/question/24372553

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6 0

11 months ago

What is the x for 1/2x=150

notsponge [240]

X=300. basically multiply 150 times 2 to get the number. To check if it’s right divided 300 by 2 and see if it’s 150 or not. If not Choose another number

6 0

2 years ago

If you were calculating the volume of a cube that has a side length of 8 inches, how would you write the calculations in exponen

Dmitry_Shevchenko [17]

Answer: 8^3, Eight to the third power, Eight times eight times eight.

6 0

3 years ago

\int (x+1)\sqrt(2x-1)dx

Nezavi [6.7K]

Answer:

$\int (x+ 1) \sqrt{2x-1} dx = \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15}(2x-1)^{\frac{5}{2}} + C$

Step-by-step explanation:

$\int (x+1)\sqrt {(2x-1)} dx\\Integrate \ using \ integration \ by\ parts \\\\u = x + 1, v'= \sqrt{2x - 1}\\\\v'= \sqrt{2x - 1}\\\\integrate \ both \ sides \\\\\int v'= \int \sqrt{2x- 1}dx\\\\v = \int ( 2x - 1)^{\frac{1}{2} } \ dx\\\\v = \frac{(2x - 1)^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}} \times \frac{1}{2}\\\\v= \frac{(2x - 1)^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{2}\\\\v = \frac{2 \times (2x - 1)^{\frac{3}{2}}}{3} \times \frac{1}{2}\\\\v = \frac{(2x - 1)^{\frac{3}{2}}}{3}$

$\int (x+1)\sqrt(2x-1)dx\\\\ = uv - \int v du$

$= (x +1 ) \cdot \frac{(2x - 1)^{\frac{3}{2}}}{3} - \int \frac{(2x - 1)^{\frac{3}{2}}}{3} dx \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ u = x + 1 => du = dx \ ]$

$= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \int (2x - 1)^{\frac{3}{2}}} dx\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \times ( \frac{(2x-1)^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}) \times \frac{1}{2}\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{3} \times ( \frac{(2x-1)^{\frac{5}{2}}}{\frac{5}{2} }) \times \frac{1}{2}\\\\= \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15} \times (2x-1)^{\frac{5}{2}} + C\\\\$

6 0

3 years ago