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BabaBlast [244]
3 years ago
13

What are the zero(s) of the function f(x)=

rmula1" title="\frac{4x^{2}-36x }{x-9}" alt="\frac{4x^{2}-36x }{x-9}" align="absmiddle" class="latex-formula">?
a.) x = -9
b.)x = 0
c.)x = 9
d.)x = 0 and x = 9
Mathematics
1 answer:
Contact [7]3 years ago
7 0

Answer:

D

Step-by-step explanation:

The function will be zero when the numerator is equal to 0. So, set the numerator equal to 0 and solve:

0 = 4 {x}^{2}  - 36x

I will use the quadratic formula:

\frac{ 36  +  \sqrt{ {36}^{2}  - 4(4)(0)} }{2(4)}  \\  \frac{  36 + 36}{8}  = 9

Therefore, x = 9 is a 0. Let's check with subtracted root now:

\frac{ 36   -    \sqrt{ {36}^{2}  - 4(4)(0)} }{2(4)}  \\  \frac{ 36  -  36}{8}  =  0

It appears 0 is also a root.

Therefore, the answer is D.

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I really need help with this problem steps
melamori03 [73]
Solve for y by setting one equation =to X after that you can substitute the equation into the other one let's set the first one in X= FORM
X-7y=10
-2x+14y=-20

Add 7y to both sides of the first equation to let x stand alone
X=7y+10
Now you can substitute x on the second equation
-2 (7y+10)+14y=-20
Distribute
-14y-20+14y=-20
Add and simplify
Us cancel out =0
-20=-20
0=0
They are the same equations you can also divide the second equation by -2 which would make it look like this
X-7y=10
-2 (x-7y=10)
Let me know if I have answered your question
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