If u is an algebraic expression and c is a real number such that c>0, then the equation |u|>c is equivalent to a. u>-c or u>c. b. u<-c or u>c. c. u<-c or u

Answers

Answer 1

Answer:

Answer: u<-c or u>c (choice B)

Explanation

Consider the example |x| > 5.

On a number line this will mean either x > 5 or x < -5. We're looking for values of x such that they are more than 5 units away from 0.

A value like x = -7 fits the description (since it's 7 units away from 0). While something like x = 1 isn't a solution.

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An equation was created for the line of best fit from the actual enrollment data. It was used to predict the dance studio enrollment values shown in the table below:Enrollment Month January February March April May June Acutal 120 140 150 140 150 130 Predicted 80 150 110 150 110 150 Residual 40 −10 40 −10 40 −20 Analyze the data. Determine whether the equation that produced the predicted values represents a good line of best fit. No, the equation is not a good fit because the residuals are all far from zero. No, the equation is not a good fit because the sum of the residuals is a large number. Yes, the equation is a good fit because the residuals are not all far from zero. Yes, the equation is a good fit because the sum of the residuals is a small number.
A student is offered two different summer jobs. Job A lasts 16 weeks and pays $350.00 per week. Job B lasts 4 months and pays $1,500.00 per month, with monthly raises of 10% of the previous month's salary. How much more does the student earn by accepting Job B (rounded to the nearest cent)?
Please i need help! first answer i will mark brainliest!
Write the number 144 as a row of prime numbers multiplied together
what is the least number that must be added to 2000so that the sum is devided exactly by 10,12,16 and 18

What is the next number in the sequence 0, 3, 8, 15, 24, 35

Answers

48
Cuz
0+3=3
3+5=8
8+7=15
15+9=24
24+11=35
35+13=48

48 think of it as increasing by n, n+2, n+4... For each number when n= an odd nimber

9. A sector of a circle has central angle $(\pi )/(4)$
and area
$(49\pi)/(8)$

${cm}^(2)$
. Find the radius of the circle

Answers

Answer:

Step-by-step explanation:

Begin with the formula for the area of a sector:

$A=(\theta)/(2\pi)*\pi r^2$ and we are given everything except the radius, r. Filling in what we know:

$(49\pi)/(8)=((\pi)/(4) )/(2\pi)*\pi r^2$ and then start the simplification process. I began by dealing with the fraction over a fraction thing, which simplifies the equation to:

$(49\pi)/(8)=(1)/(8)\pi r^2$ then isolate the r-squared:

$(8(49\pi))/(8\pi)=r^2$ . Canceling out the π's and the 8's leaves us with simply

r² = 49 so

r = 7

How do you factor k^2-k-56

Answers

(k-8)(k+7) is how you factor is

use the factoring method
(k-8)(k+7)

Simplity
4x2 – 3x + 7 + 2x2 + 4x

Answers

Answer:

6x^2+x+7

Step-by-step explanation:

4x^2-3x+7+2x^2+4x

4x^2+2x^2+4x-3x+7

6x^2+x+7

Which of the sets of ordered pairs represents a function?A = {(1, −2), (3, −5), (5, 2), (7, 5)}

B = {(4, 2), (4, −2), (9, 3), (9, −3)}

only A
only B
Both A and B
neither A or B

Answers

Answer:

Option A is correct.

only set A is the ordered pairs represents a function.

Step-by-step explanation:

By the definition of a function, for each input there exists a unique output.

In particular, for each value of x there exists unique value of y.

In set A,

we have four ordered pair (1, -2) , (3, -5) , (5, 2) and (7, 5)

For each input value of x in set A, there is a unique value of y in set A.

therefore, Set A is a function.

In set B,

we have four ordered pair (4, 2) , (4, -2), (9, 3) and (9, -3)

which means for x=4 , we have two values of y, i.,e 2 and -2

⇒ for a single value of x , there exist more than one value of y,

therefore set B is not a function

It's only A, because for each x, there's only one y, unlike B where there are two values for x=4 and two for x=9.

Find the volume of a cube whose side length is 5x^2

Answers

V = s³
V = (5x²)³
V = (5x²)(5x²)(5x²)
V = 5x²+²+²
V = 5x^6

If the side length is 5 x 2, 5 x 2 = 10, so 10 x 3 is 30. Hope this helped!

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