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Sonja [21]
3 years ago
7

What does 4x+7x equal ?

Mathematics
2 answers:
andrew11 [14]3 years ago
8 0
11x :) your welcomen
olganol [36]3 years ago
7 0

Answer:

11x

Step-by-step explanation:

u just add them regularly

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What's The Answers To Both Of These Problem's
konstantin123 [22]
Mathway .com gives you FREE answers (its an app too)<span>. you could also download Photomath app </span>for FREE! read more about in the app descriptions :)

7 0
3 years ago
(?,5) is on the line 4x-7y=1. Find the other half of the coordinate.
ludmilkaskok [199]
The coordinate is an ordered pair (x,y). We have a number, 5, for y, which we can substitute (plug-in) for y in the equation of the line and solve for x.

4x-7y=1\\4x-7(5)=1\\4x=36\\x=9

(9,5)
8 0
3 years ago
Perform the indicated operation. Simplify the result in factored form. ((x - y)/ (x^2 - 1)) * ((x - 1)/(x^2 - y^2))
Scilla [17]

Answer:

The answer to your question is            \frac{1}{(x + 1)(x + y)}            

Step-by-step explanation:

Real operation

                      \frac{x - y}{x^{2} - 1} \frac{x - 1}{x^{2} - y^{2}}

Process

1.- Factor the denominators

                     \frac{x - y}{(x + 1)(x - 1)} \frac{x - 1}{(x + y)(x - y)}

2.- Simplify like terms

Cancel (x - y) and (x - 1) because they are in numerator and the denominator

                    \frac{1}{(x + 1)(x + y)}            This is the answer or you can expand

3.- Expand

                   \frac{1}{x^{2} + xy + x + y^{2}}                    

3 0
3 years ago
What is an equivalent form of the function f(x) = x^2 +8x+15 that reveals the zeros of the function? A. y= (x-3)^2-5 B. f(x) = (
777dan777 [17]

ANSWER

C. f(x) = (x+3)(x+5)

EXPLANATION

The given function is

f(x) = x^2 +8x+15

The factored form of the function reveals the zeros of the function.

From the factored form, we apply the zero product principle to get the zeros.

From the given options, the factored form of the polynomial is

f(x) = (x+3)(x+5)

8 0
3 years ago
Solve for x:
Marta_Voda [28]
The answer is C.) x=15
7 0
3 years ago
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