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Ganezh [65]
3 years ago
7

How to use the zero product property and factoring solve for X

Mathematics
1 answer:
Marta_Voda [28]3 years ago
8 0
Using the zero product property you can set each expression equal to 0 (or factor for some of them)
All you have to do is set them equal to zero and simplify to solve for x

a. (x+13)=0
x= -13
(x-7)=0
x=7

b. (2x+3)=0
x=-3/2
(3x-7)=0
x=7/3

c. x=0
(x-3)=0
x=3

d. x^{2} -5x=0
 needs to be factored 
factor out an x to get:
x(x-5) and set each equal to 0
x = 0       
(x-5)=0 
x=5

e. x^{2} -2x-35=0
 also needs to be factored
(x-7)(x+5)=0
(x-7)=0       (x+5)=0
x=7   x=-5

f. 3x^{2} +14x-5=0
Also needs to be factored using the ac method:
3 x^{2} +15x-x-5
3 x^{2} (x+5)-1(x+5)
(3x-1)(x+5)=0

(3x-1)=0
x=1/3
(x+5)=0
x=-5

I hope this helps! (if so, give brainliest please!!)

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ladessa [460]

Answer:

lol at the pictures it has all the work

7 0
3 years ago
Look at the figure. Which ray is opposite of DE?
exis [7]

Answer:

BA

Step-by-step explanation:

Its exactly opposite of DE.

3 0
4 years ago
Someone Help Please:))
timama [110]

Answer:

(0,3), (1,4) (2,5) and (3,6)

Step-by-step explanation:

f(x) = x + 3 is equivalent to y = 1x + 3 (y = mx + b)

b = the y intercept, when x is 0.

So y = 3, which means its (0,3)

To check:

f(x) = x + 3

f(0) = 0 +3

f(0) = 3

(0,3)

Now let's try that for all the values (1,2 and 3)

f(x) = x + 3

f(1) = 1 + 3

f(1) = 4

(1,4)

f(x) = x + 3

f(2) = 2 + 3

f(2) = 5

(2,5)

f(x) = x + 3

f(3) = 3 + 3

f(3) = 6

(3,6)

Hope that helps!

4 0
2 years ago
Ella has 50 stacks of 10 pennies in each sack describe how to find how many pennies Ella had in all
GrogVix [38]
She has 50 stacks with 10 pennies in each, so multiply 50 by 10
50*10 = 500
so she has 500 pennies in total.
7 0
3 years ago
Read 2 more answers
The expression 2x−(2x+3)2+4(x−1)+x3 can be simplified into the form Ax3+Bx2+Cx+D. Find the missing constants A, B, C, and D belo
Talja [164]

Answer:

Step-by-step explanation:

Before solving this we  have to know that,

(-) ×(-)=(+)

(+)×(+) =(+)

(+)×(-)=(-)

Lets solve now,

2x-(2x+3)2+4(x-1)+x^{3} \\2x-(2x-3)^{2} +4x-4+x^{3} \\2x-(4x^{2} -12x+9)+4x-4+x^{3} \\2x-4x^{2} +12x-9+4x-4+x^{3} \\2x+12x+4x-4x^{2} -9-4+x^{3} \\14x-4x^{2} -13+x^{3} \\\\So,\\x^{3} -4x^{2} +14x-13\\\\Therefore,\\A = 1\\B = (-4)\\C = 14\\D = (-13)

Hope this helps you

Let me know if you have any other questions :-)

4 0
3 years ago
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