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
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2, factor out the leading coefinet (constant in front of the x^2 term), basicallly factor out a
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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
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4. complete the squre and expand
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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
vertex is (8,-3)
Answer:
n=64 and n+1=65
Step-by-step explanation:
64 + 65 equals 129
Answer:
Step-by-step explanation:
-3x^2 --2x--2 -- 9/x--2
Answer:
a = 2x
b = 3
Step-by-step explanation:
Consider the sum of cubes identity
a³ + b³ =(a + b)(a² -ab +b²)
for the polynomial 8x³ +27, we factorise
8x³ + 27
∛8 = 2
∛8 = x
∛27 = 3
Therefore, we can say that :
8x³ + 27 = (2x)³ + 3³
Using this: a³ + b³ =(a + b)(a² -ab +b²)
We can say that
a = 2x
b = 3
To confirm that: a = 2x and b = 3 we factorise 8x³ + 27
= (2x)³ + 3³
= (2x + 3)((2x)²− 2x × 3 + 3²)
= (2x + 3)(4x² - 6x + 9)
= 8x³ - 12x² +18x + 12x² - 18x + 27
= 8x³ + 27
Therefore, a = 2x and b = 3 is correct.
=
Hi
16.8/0.8 = 168/8 = 21
168 |__8__
-16 21
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08
- 08
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Answer: 21
