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3 years ago

In the accompanying diagram, parallelogram ABCD has

Mathematics

1 answer:

Taya2010 [7]3 years ago

5 0

Answer:

The area of the paralellogram is 24 square units.

Step-by-step explanation:

Geometrically speaking, the area of the parallelogram has an equation equivalent to the area formula for a rectangle, that is:

$A = b\cdot h$ (1)

Where:

$A$ - Area.

$b$ - Base.

$h$ - Height.

The base and height of the parallelogram are, respectively:

Base

$b = 8-2$

$b = 6$

Height

$h = 5-1$

$h = 4$

Then, the area of the parallelogram is:

$A = (6)\cdot (4)$

$A = 24$

The area of the paralellogram is 24 square units.

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Answer:

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Step-by-step explanation:

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3 years ago

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brilliants [131]

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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