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3 years ago

Solve the following equation: 4(x + 2) - 17 = 35 A: -11 B: 25/2 C: 15 D: 11

Mathematics

2 answers:

aalyn [17]3 years ago

5 0

Answer:

D: 11

Step-by-step explanation:

4(x + 2) - 17 = 35

4x+8 - 17=35

4x-9=35

4x=35+9=44

x=44/4=11

Rzqust [24]3 years ago

4 0

The answer is D, the working out is in the photo

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Step-by-step explanation:

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3 years ago

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3 years ago

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3 years ago

The Davis family is going to reseed the grass in the backyard. Their yard is irregularly shaped with a cement patio and rectangu

prohojiy [21]

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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. $\square$

7 0

2 years ago