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il63 [147K]
3 years ago
15

Simplify the following: 3x+2y+7-5x+7y

Mathematics
2 answers:
const2013 [10]3 years ago
3 0
<span>3x+ 2y+7 -5x+7y
= -2x + 9y + 7

hope it helps</span>
WARRIOR [948]3 years ago
3 0
9y-2x+7   

3x-5x= -2x

2y+7y=9y                                                                                                                                                                       
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24w²-16 <br><br> I need help factoring out the greatest common factor of the polynomial
mestny [16]

Hello from MrBillDoesMath!

Answer:

8 (3w^2 - 2)


Discussion:

The  greatest common factor is 8, so

24w^2 - 16 = 8 (3w^2 - 2)


Thank you,

MrB

6 0
3 years ago
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Which two equations would be most appropriately solved by using the zero product property? Select each correct answer.
LekaFEV [45]
The zero product property tells us that if the product of two or more factors is zero, then each one of these factors CAN be zero.

For more context let's look at the first equation in the problem that we can apply this to: (x-3)(x+4)=0

Through zero property we know that the factor (x-3) can be equal to zero as well as (x+4). This is because, even if only one of them is zero, the product will immediately be zero.

The zero product property is best applied to factorable quadratic equations in this case.

Another factorable equation would be 2x^{2}+6x=0 since we can factor out 2x and end up with 2x(x+3)=0. Now we'll end up with two factors, 2x and (x+3), which we can apply the zero product property to.

The rest of the options are not factorable thus the zero product property won't apply to them.
3 0
3 years ago
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Can someone explain how to do 1-3. I don’t understand how to do them :(
sattari [20]

Answer:

-2

Step-by-step explanation:

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3 0
3 years ago
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4)Without a calculator, name the zeros and sketch the graph.<br> 56.) = -2x(x-4)^2(x+3)^2
Ilia_Sergeevich [38]

Answer:

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Step-by-step explanation:

replace y with 0 and then fro the sketch I recommend you tomuse desmos good luck buddie

6 0
3 years ago
H(j + i) - k5 + h<br> when h = 3, i = 4, j = 2, and k = 1<br> pls put how you did it
VikaD [51]

Answer:

16

Step-by-step explanation:

First you want to plug in all the numbers where the letters are

h ( j + i )- k(5) + h

3 ( 2 + 4 )- 1(5) + 3

After this step you will have to distribute the 3 outside the parenthesis. Which you do by multiplying everything in side the parenthesis by the outside number which is three.

6 + 12 - 1(5) + 3

Then you just follow PEMDAS to solve for the rest

6 + 12 - 5 + 3

18 - 5 + 3

13 + 3

16

6 0
3 years ago
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