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3 years ago

Rewrite the following without an exponent

Mathematics

1 answer:

Viefleur [7K]3 years ago

5 0

The fraction 8/5 is raised to the power -1. If a fraction is raised to a negative power, its reciprocal will be having the positive power.

For example in case of $(x)^{-1}$ , it can be written as $\frac{1}{x}$ . i.e. to remove the exponent -1 we simply take the reciprocal.

So for given expression the reciprocal will be 5/8

Therefore,

$( \frac{8}{5} )^{-1}= \frac{5}{8}$

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The verticles of a polygon p(0,4), q(5,4), r(5.-3) and s(0,-3). What are the perimeters of the polygon

makkiz [27]

Answer:

Following are the responses to the given question:

Perimeter of polygon:

$PQ = 5\\\\RS = 5\\\\PR = QS=\sqrt{74} = 8.60\\\\QR = 7$

Step-by-step explanation:

In this question we calculate the perimeter of a polygon which can be defined as follows:

Using formula:

$d = \sqrt{(x _2 - x_1)^2 + (y_2 - y_1)^2}$

When vertices are:

$P ( 0 , 4) \\\\Q(5 , 4)$

$x_1 = 0\\\\y_1 = 4\\\\x_2 = 5\\\\y_2 = 4\\\\PQ = \sqrt{( 5 - 0)^2 + ( 4 - 4)^2}\\\\= \sqrt{5^2 + 0^2}\\\\= \sqrt{25}\\\\= 5$

$R( 5 , -3) and S ( 0 , -3)x_1 = 5y_1 = -3x_2 = 0\\y_2 = -3\\RS = \sqrt{ ( 0 - 5)^2 + ( -3 - (-3))^2}\\= \sqrt{( -5)^2 + ( -3 + 3)^2}\\= \sqrt{ 25 + 0}\\= \sqrt{25}\\= 5\\P( 0 , 4)\ and\ R (5 , -3)\\x_1 = 0\\y_1 = 4\\x_2 = 5\\y_2 = -3\\PR = \sqrt{( 5 - 0)^2 + ( -3 - 4)^2}\\= \sqrt{5^2 + (-7)^2}\\= \sqrt{ 25 + 49}\\= \sqrt{74}\\= 8.60\\$

$Q ( 5 , 4) \ and \ S(0 ,-3)\\x_1 = 5\\y_1 = 4\\x_2 = 0\\y_2 = -3\\QS =\sqrt{ ( 0 - 5)^2 + ( -3 - 4)^2}\\= \sqrt{-5^2 + -7^2}\\= \sqrt{25 + 49}\\= \sqrt{74}\\= 8.60$

$Q( 5,4\ and\ R(5 , -3)\\x_1 = 5\\y_1 = 4\\x_2 = 5\\y_2 = -3\\QR = \sqrt{(5 - 5)^2 + ( -3 - 4)^2}\\= \sqrt{0^2 + (-7)^2}\\= \sqrt{0 + 49}\\= \sqrt{49}\\= 7\\$

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Step-by-step explanation:

just did on edg.

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