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gladu [14]
2 years ago
11

Determine which polynomial has (x + 4) as a factor.

Mathematics
1 answer:
Amanda [17]2 years ago
5 0

Step-by-step explanation:

→ Option A

x² - 11x + 28

= (x - 7)(x - 4) ❎

→ Option B

x² - 6x + 8

= (x - 4)(x - 2) ❎

→ Option C

x² + 7x + 12

= <u>(x + 4)</u>(x + 3) ✔️

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The equation is ax+r=7 <br><br>how do I solve for r??
Strike441 [17]
The solution is r= 7/a-x How did i do this?
Your equation is ax+ r=7, so the goal is to isolate the r. You begin by dividing the a from the ax. Remember, what you do to one side, you must do to the other. So you do 7/a, since you can't simplify 7/a, leave it as it is. Next, your equation should look like this x+r=7/a. In order to isolate the r, you move the x to the other side by subtracting, which leaves you with the answer of r= 7/a-x.
Hope this helped!
7 0
3 years ago
Read 2 more answers
The expression 4x/5-2x/3 is equivalent to what?
Vinvika [58]

Answer:

(2 x)/15

Step-by-step explanation:

Simplify the following:

(4 x)/5 - (2 x)/3

Put each term in (4 x)/5 - (2 x)/3 over the common denominator 15: (4 x)/5 - (2 x)/3 = (12 x)/15 - (10 x)/15:

(12 x)/15 - (10 x)/15

(12 x)/15 - (10 x)/15 = (12 x - 10 x)/15:

(12 x - 10 x)/15

12 x - 10 x = 2 x:

Answer: (2 x)/15

7 0
3 years ago
B) A monthly service fee to download e-books is $4.95, plus $3.99 per book downloaded. Write an
pickupchik [31]

Answer:

$4.95 + $3.99b = c

I can't make a graph but start at 4.95 on the graph and move up 3.99 every time you go to the right once

7 0
3 years ago
How do I simplify the expression 2-^1/3 times 2^7/3<br> A. 64<br> B. 1/4<br> C. 4<br> D. 8
mihalych1998 [28]

Answer:

C.4

Step-by-step explanation:

Since the values are of the same base....

we add the powers..because its multiplication...

;2^(-1/3 + 7/3)

;2^(6/3) = 2^2

;Ans = 2^2 = 4

8 0
2 years ago
a family walked 240 miles in 4 hours yesterday at this rate how many hours will it take to walk another 150 miles ​
alekssr [168]

Answer:

2.5 hours. my bed if it wrong

Step-by-step explanation:

3 0
2 years ago
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