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3 years ago

Which mixed number is equivalent to the improper fraction? 21/9

Mathematics

1 answer:

algol133 years ago

5 0

Answer:

2 1/3

Step-by-step explanation:

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15; it represents the one-time sign-up fee

The y-intercept can be found either on the graph where the line intercepts the y-axis, or b in the equation y=mx+b.

It represents the fee because it will be charged even if the number of months (x) is zero. 10 is the monthly fee because it is multiplied by x, the number of months.

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Simplify the expression. Assume that all variables represent nonzero real numbers.StartFraction (4 n Superscript 4 Baseline q Su

Alja [10]

Answer:

$\frac{ - 3}{ 256 {q}^{10} {n}^{8} }$

Step by step explanation:

$\frac{ {(4 {n}^{4} {q}^{5})}^{2} {(8 {n}^{4} q)}^{-2} }{ {(- 3 {nq}^{9})}^{ - 1} {(4 {n}^{3} {q}^{9}) }^{3} }$

first we will change the terms with negative superscrips to the other side of the fraction

$\frac{{(4 {n}^{4} {q}^{5})}^{2}{(- 3 {nq}^{9})}^{ 1}}{{(4 {n}^{3} {q}^{9})}^{3} {(8 {n}^{4} q)}^{2} }$

then we will distribute the superscripts

$\frac{ {4}^{2} {n}^{2 \times 4} {q}^{2 \times 5} (- 3) {nq}^{9}}{ {4 }^{3}{n}^{3 \times 3} {q}^{9 \times 3} {8 }^{2}{n}^{4 \times 2} {q}^{2} }$

$\frac{ {4}^{2} {n}^{8} {q}^{10} (- 3) {nq}^{9}}{ {4 }^{3}{n}^{9} {q}^{27} {8 }^{2}{n}^{8} {q}^{2} }$

as when multiplying two powers that have the same base, we can add the exponents and, to divide podes with the same base, we can subtract the exponents

${4}^{2 - 3} {q}^{10 + 9 - 2 - 27} {n}^{8 + 1 - 8 - 9} {8}^{ - 2} { (- 3)}^{1}$

${4}^{ - 1} {q}^{ - 10} {n}^{ - 8} {8}^{ - 2} { (- 3)}^{1}$

then we will change again the terms with negative superscrips to the other side of the fraction

$\frac{ - 3}{ 4 \times {8}^{2} {q}^{10} {n}^{8} }$

$\frac{ - 3}{ 256 {q}^{10} {n}^{8} }$

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3 years ago