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Bogdan [553]
3 years ago
7

Is (8,8) a solution to the equation y=x?​

Mathematics
2 answers:
Nookie1986 [14]3 years ago
7 0

Answer:

Yes

Step-by-step explanation:

(8,8) is a solution is y=x.

If we substitute the point into the equation, the equation is true.

(8,8) is in (x,y) format

Substitute:

y=x

8=8

LS=RS (left side equal rights side)

Mnenie [13.5K]3 years ago
7 0

Answer:

yes

Step-by-step explanation:

y=x is just a linear function and the slope is 1 to 1.

so the y-intercept is 0 and the the line goes (1,1),(2,2),(3,3),(4,4).....etc so as x increases by 1, y increases by 1 and if it

makes it easier look at this:

y=1x+0

now look at this:

y=x

aren't they the same?

Hope that helps.

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EXPLANATION

The volume of a sphere is given by the following relationship:

\text{Volume}=\frac{4}{3}\cdot\pi\cdot(r)^3

Replacing terms:

\text{Volume = }\frac{4}{3}\cdot\pi\cdot(10)^3\text{Volume = }\frac{4}{3}\cdot3.14\cdot1000

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\text{Volume = 4186.67 yd}^3

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For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[
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Answer:

The following are the solution to the given points:

Step-by-step explanation:

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1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

Solve point 1 that is \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\:

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k= 1 \to  s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\

                  = \frac{1}{2} - \frac{1}{3}\\\\

k= 2 \to  s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\

                  = \frac{1}{3} - \frac{1}{4}\\\\

k= 3 \to  s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\

                  = \frac{1}{4} - \frac{1}{5}\\\\

k= n^  \to  s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\

Calculate the sum (S=s_1+s_2+s_3+......+s_n)

S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\

   =\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\

When s_n \ \ dt_{n \to 0}

=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\

\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}

In point 2: \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}

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k= 1 \to  s_1 = \frac{1}{(1+6)(1+7)}\\\\

                  = \frac{1}{7 \times 8}\\\\= \frac{1}{56}

k= 2 \to  s_1 = \frac{1}{(2+6)(2+7)}\\\\

                  = \frac{1}{8 \times 9}\\\\= \frac{1}{72}

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                  = \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\

k= n^  \to  s_n = \frac{1}{(n+6)(n+7)}\\\\

calculate the sum:S= s_1+s_2+s_3+s_n\\

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\

when s_n \ \ dt_{n \to 0}

S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066

\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}

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