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Pavlova-9 [17]
3 years ago
11

Solve for x: 3 < x + 3 < 6 06 06 >x>9 00 00>x>3

Mathematics
1 answer:
alexandr1967 [171]3 years ago
6 0

Answer:

6<0 is the right answer

Step-by-step explanation:

this is due to addition

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NEED HELP ASAP!
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complette the square to get vertex form or y=a(x-h)^2+k
(h,k) is vertex
1. group x terms, so for y=ax^2+bx+c, do y=(ax^2+bx)+c <span> <span> </span> 2, factor out the leading coefinet (constant in front of the x^2 term), basicallly factor out a </span><span> <span> </span> 3. take 1/2 of the linear coefient (number in front of the x), and square it ,then add negative and positive of it inside parnthases </span><span> <span> </span> 4. complete the squre and expand </span>


so
y=-1/4x^2+4x-19
group
y=(-1/4x^2+4x)-19
undistribute -1/4
y=-1/4(x^2-16x)-19
take 1/2 of -16 and squer it to get 64 then add neg and pos inside
y=-1/4(x^2-16x+64-64)-19
factorperfect square
y=-1/4((x-8)^2-64)-19
expand
y=-1/4(x-8)^2+16-19
y=-1/4(x-8)^2-3
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