PLEASE HELP OMG PLEASE DUE TOMORROW AND MY FINAL GRADE SO PLEASE JUST HELP.

Mathematics

1 answer:

uranmaximum [27]2 years ago

3 0

The second answer seems to be correct. Hope this helps :)

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Which number can each term of the equation be multiplied by to eliminate the decimals before solving?

Sati [7]

Ok...

5.6 = 56/10 = 560/100

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--------------

$\frac { 560 }{ 100 } j-\frac { 12 }{ 100 } =4+\frac { 110 }{ 100 } j\\ \\ \\ 100\times \left( \frac { 560 }{ 100 } j-\frac { 12 }{ 100 } \right) =\left( 4+\frac { 110 }{ 100 } j \right) \times 100\\ \\ 560j-12=400+110j\\ \\ 560j-110j=400+12\\ \\ 450j=412\\ \\ j=\frac { 412 }{ 450 }$

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3 years ago

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Please please help me

Gekata [30.6K]

Answer:

$\large\boxed{\dfrac{\boxed{8}}{81}}$

Step-by-step explanation:

$\text{If}\ a_1,\ a_2,\ a_3,\ a_4,\ ...,\ a_n\ \text{is the geometric sequence, then}\\\\\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...=\dfrac{a_n}{a_{n-1}}=constant=r-\text{common ration}.\\\\\text{We have}\ \dfrac{1}{2},\ \dfrac{1}{3},\ \dfrac{2}{9},\ \dfrac{4}{27},\ ...\\\\\dfrac{\frac{1}{3}}{\frac{1}{2}}=\dfrac{1}{3}\cdot\dfrac{2}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{2}{9}}{\frac{1}{3}}=\dfrac{2}{9}\cdot\dfrac{3}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{4}{27}}{\frac{2}{9}}=\dfrac{4}{27}\cdot\dfrac{9}{2}=\dfrac{2}{3}$

$\bold{CORRECT}\\\\\dfrac{x}{\frac{4}{27}}=\dfrac{2}{3}\qquad\text{multiply both sides by}\ \dfrac{4}{27}\\\\x=\dfrac{2}{3}\cdot\dfrac{4}{27}\\\\x=\dfrac{8}{81}$

8 0

2 years ago