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3 years ago

HELP ME AND GIVE ME THE ANSWER I DONT NEED U TO EXPLAIN ANYTHING JUST HURRY UP

Mathematics

1 answer:

Brums [2.3K]3 years ago

6 0

Answer:

x = 9.1

Step-by-step explanation:

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Karo-lina-s [1.5K]

Answer:

32t+12

Step-by-step explanation:

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3 years ago

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How do I find the mean of these numbers

pshichka [43]

Answer:

add all of them together (x and y’s) the divide the sum by the amt of numbers

Step-by-step explanation:

8 0

2 years ago

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Which answers are equivalent to ? Choose all answers that are correct. A. 0.08 B. 8% C. 0.32 D. 32%

Llana [10]

A and B are equivalent and C and D are equivalent

4 0

4 years ago

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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

How to do quadratic graphs

stealth61 [152]

Answer:

To graph a quadratic equation, start by solving for h in vertex form, or taking -b divided by 2 times a in standard form. Then, define or calculate the value of k and plot the point (h, k), which is the vertex of your parabola.

Step-by-step explanation:

7 0

4 years ago