All categories

3 years ago

What is a muscle verb??

2 answers:

8 0

Definition of muscle (Entry 2 of 2) transitive verb. : to move or force by or as if by muscular effort muscled him out of office. intransitive verb. : to make one's way by brute strength or by force.

Rashid [163]3 years ago

8 0

Answer: move (an object) in a particular direction by using one's physical strength. -angie:) . pls mark me brainliest! have a nice day!

Step-by-step explanation: pls trust me

You might be interested in

1. Find the Surface Area of the rectangular prism. Show your work and explain how you solved using First,

mafiozo [28]

Answer:

376 inches^2

Step-by-step explanation:

<h3>1. Divide up each type of rectangle </h3>

- 10x6

- 6x8

- 8x10

<h3>2. Find the area of each rectangle</h3>

- 10x6=60

- 6x8=48

- 8x10=80

<h3>3. Multiply each rectangle by two to get the opposite one</h3>

- 60x2=120

- 48x2=96

- 80x2=160

<h3>4. Add them together</h3>

120+96+160=376

The surface area is 376 inches^2

6 0

3 years ago

3 0.075, 0.75%, 4 20. Which of the following correctly orders the values above from least to greatest?

Dafna1 [17]

0.075, 0.75, 3, 4, 20

8 0

3 years ago

Read 2 more answers

Write another name for < 5

victus00 [196]

Answer:

less than 5

x < 5

6 0

3 years ago

Read 2 more answers

Given LaTeX: f\left(x\right)=x^{^3}-3x+4f ( x ) = x 3 − 3 x + 4, determine the intervals where the function is increasing and wh

Vedmedyk [2.9K]

Answer:

Increasing: $x$ and $x>1$ .

Decreasing: $-1$

Step-by-step explanation:

We have been given a function $f(x)=x^3-3x+4$ . We are asked to determine the intervals, where the function is increasing and where it is decreasing.

First of all, we will find critical points of our given function by equating derivative of our given function to 0.

Let us find derivative of our given function.

$f'(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x)+\frac{d}{dx}(4)$

$f'(x)=3x^{3-1}-3+0$

$f'(x)=3x^{2}-3$

Let us equate derivative with 0 as find critical points as:

$0=3x^{2}-3$

$3x^{2}=3$

Divide both sides by 3:

$x^{2}=1$

Now we will take square-root of both sides as:

$\sqrt{x^{2}}=\pm\sqrt{1}$

$x=\pm 1$

$x=-1,1$

We know that these critical points will divide number line into three intervals. One from negative infinity to -1, 2nd -1 to 1 and 3rd 1 to positive infinity.

Now we will check one number from each interval. If derivative of the point is greater than 0, then function is increasing, if derivative of the point is less than 0, then function is decreasing.

We will check -2 from our 1st interval.

$f'(-2)=3(-2)^{2}-3=3(4)-3=12-3=9$