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3 years ago

Which statement is true concerning the vertex and axis of symmetry of h(x)=-2x^2+8x?

Mathematics

2 answers:

ludmilkaskok [199]3 years ago

6 0

My favorites yumAnswer:

Step-by-step explanation:

im jk the answer is 6 bc u multiply them and divide by 8 u would then divide that by 2

stich3 [128]3 years ago

5 0

Answer: the vertex is at (2, 8) and the axis of symmetry is x = 2

Step-by-step explanation:

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3 years ago

How do I solve 3w+6=4w ?

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4 years ago

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<img src="https://tex.z-dn.net/?f=%5Csqrt%7B%28-3-4%29%5E%7B2%7D%2B%28-7-3%29%5E%7B2%29%20%7D" id="TexFormula1" title="\sqrt{(-3

bulgar [2K]

Answer:

Step-by-step explanation:

$\sqrt{(-3-4)^2+(-7-3)^2$

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3 years ago

Use mathematical induction to prove

Alex17521 [72]

$Prove\ that\ the\ assumption \is \true for\ n=1\\1^3=\frac{1^2(1+1)^2}{4}\\ 1=\frac{4}{4}=1\\$

Formula works when n=1

Assume the formula also works, when n=k.

Prove that the formula works, when n=k+1

$1^3+2^3+3^3...+k^3+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k+1)^2}{4}+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k^2+2k+1)}{4}+(k+1)^3=\frac{(k^2+2k+1)(k^2+4k+4)}{4} \\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+4k^3+4k^2+2k^3+8k^2+8k+k^2+4k+4}{4}\\\\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+2k^3+k^2}{4}+\frac{4k^3+12k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+6k^3+13k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\$

Since the formula has been proven with n=1 and n=k+1, it is true. $\square$

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2 years ago