3 years ago

PLEASE ANSWER THIS AND IF YOU DO YOU GET 1. FOLLOW 2. BRAINLIEST

Mathematics

1 answer:

snow_lady [41]3 years ago

7 0

First: 0
Second: 0
Third: -5
Fourth: y=4x

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What is one third of 27 over 48

blsea [12.9K]

27 divided by 3 is 9
48 divided by 3 is 16
9/16

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3 years ago

P=3h+3u solve for h<br> ...?

horrorfan [7]

p=3h + 3u subtract 3u from both sides

p-3u=3h now divide both sides by 3

h = p/3 - u

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4 years ago

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What is the product of <img src="https://tex.z-dn.net/?f=%283y%20%5E%7B-4%7D%29%282y%5E%7B-4%7D%20%29" id="TexFormula1" title="(

ludmilkaskok [199]

Answer:

final answer is 6/(y^8)

Step-by-step explanation:

(3y^-4) = 3/(y^4)

(2y^-4) = 2/(y^4)

3*2 = 6

Add exponents y^4+y^4 = y^8

hence 6 / (y^8)

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4 years ago

The volume, V, of any cube with side length, s, can be determined using the formula V = s3 . What is the volume, in cubic centim

storchak [24]

The answer for your question is c

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3 years ago

Use induction to show that 12 + 22 + 32 + ... + n2 = n(n+1)(2n+1)/6, for all n > 1.

dlinn [17]

Answer with Step-by-step explanation:

Let $P(n)=1^2+2^2+3^2+.....+n^2=\frac{n(n+1)(2n+1)}{6}$

Substitute n=2

Then $P(2)=1+2^2=5$

$P(2)=\frac{2(2+1)(4+1)}{6}=5$

Hence, P(n) is true for n=2

Suppose that P(n) is true for n=k >1

$P(k)=1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$

Now, we shall prove that p(n) is true for n=k+1

$P(k+1)=1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS

$P(k+1)=1^2+2^2+3^2+.....+k^2+(k+1)^2$

Substitute the value of P(k)

$P(k+)=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$

$P(k+1)=(k+1)(\frac{k(2k+1}{6})+k+1)$

$P(k+1)=(k+1)(\frac{2k^2+k+6k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+7k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+4k+3k+6}{6})$

$P(k+1)=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS=RHS

Hence, P(n) is true for all n >1.

Hence, proved

4 0

3 years ago