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3 years ago

2/x^2 in indices form explanation. I know the answer is 2x^-2 but I want an explanation.

Mathematics

1 answer:

Misha Larkins [42]3 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

Using the rule of exponents

$\frac{1}{a^{m} }$ ⇔ $a^{-m}$ , then

$\frac{2}{x^2}$ = 2 × $\frac{1}{x^2}$ = 2 $x^{-2}$

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3 years ago

The radius of a circle is 13 ft. Find its circumference in terms of \piπ.

Law Incorporation [45]

Answer:

The circumference of the circle is

Formula: 2πr

Radius=13ft

So, 2 x 22/27 x 13 = 81.71ft

Answer: 81.71ft

*Answers may differ a bit in decimals with 3.14 and direct π .

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3 years ago

Entomologists have discovered that a linear relationship exists between the rate of chirping of crickets of a certain species an

vekshin1

Answer:

a) $y=4x-160$

b) $y=4(102)-160=248\frac{chirps}{min}$

Step-by-step explanation:

We know that theres a linear relationship between the rate of the chirping of crickets and the air temperature.

The equation of a line is:

$y=mx+b$

So, let's name our variables

x= Temperature

y=Rate of the chirping

First of all, we need to find the slope with the two given points

$x_{1}=$ 60ºF , $y_{1} =80\frac{chirps}{min}$

$x_{2} =$ 80ºF, $y_{2} =160\frac{chirps}{min}$

By,

$m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} } =\frac{160-80}{80-60}$

$m=4$

Now, the equation between the air temperature and the number of chirps is:

$y-y_{1} =m(x-x_{1} )$

$y-80=4(x-60)$

Solving for y,

a) $y=4x-160$

b) To calculate the rate at which the crickets chirp when the temperature is 102 ºF we need to evaluate y(102)

$y=4(102)-160=248\frac{chirps}{min}$

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3 years ago

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Thousandths and the other one is under the hundredths

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3 years ago

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elena55 [62]

It is 25% because it is 1/4 which is a quarter.

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4 years ago