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Zigmanuir [339]

2 years ago

15

Does 8 + 2(x−6) = −2 + 2x −2 have one solution, no solutions, or infinite solutions.

2 answers:

tia_tia [17]2 years ago

7 0

Answer:

Infinite solutions

Step-by-step explanation:

All solutions are possible.

bearhunter [10]2 years ago

4 0

Answer:

answer no solution and it's attached to the pic so see and understand.

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katrin [286]

Not Rationally, No. But You Can Make It Into 26.5

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2 years ago

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11 students went on a trip to New York City. 3 of them visited the Statue of Liberty. What fraction of the students visited the

Ne4ueva [31]

so since only 3 out of 11 went to visit then it is simple

since just 3 went and there were 11

3/11 of the students went

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3 years ago

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36√(3)-(432/7+4√(3))=?

Elden [556K]

40√3+432/7 is the answer

Hope this helps :D

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2 years ago

Consider the following sequence

wlad13 [49]

Answer:

a₇ = 2.375

Step-by-step explanation:

There is a common ratio r between consecutive terms, that is

r = $\frac{-76}{152}$ = $\frac{38}{-76}$ = $\frac{-19}{38}$ = - $\frac{1}{2}$

This indicates the sequence is geometric with nth term

$a_{n}$ = a₁ $(r)^{n-1}$

where a₁ is the first term and r the common ratio

Here a₁ = 152 and r = - $\frac{1}{2}$ , then

$a_{n}$ = 152 $(-\frac{1}{2}) ^{n-1}$ , so

a₇ = 152 $(-\frac{1}{2}) ^{6}$ = 152 × $\frac{1}{64}$ = 2.375

3 0

2 years ago

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Please help solve this system of equations

stepan [7]

Make a substitution:

$\begin{cases}u=2x+y\\v=2x-y\end{cases}$

Then the system becomes

$\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}$

Simplifying the equations gives

$\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}$

which is to say,

$\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}$

$\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81$

$\implies\dfrac uv=\pm27$

$\implies u=\pm27v$

Substituting this into the new system gives

$\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1$

$\implies u=\pm27$

Then

$\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13$

(meaning two solutions are (7, 13) and (-7, -13))

8 0

2 years ago