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3 years ago

THIS IS URGENT I NEED HELP ASAP.(9x 10^-3)^2

Mathematics

1 answer:

joja [24]3 years ago

8 0

(9x 10^-3)^2= $\frac{81x^{2}}{1000000}$ , so d 8.1x10^-6

Evaluate 1/2a^-4b^2 for a =-2 and b=4

$\frac{1}{2} (-2)^{-4} *4^{2}\\\\\frac{1}{2} *\frac{1}{(-2)^{4}} *4^{2}\\\\\frac{1}{2} *\frac{1}{2^{4}}*4^{2}\\\\\frac{1}{2} *\frac{1}{16}*4^{2}\\\\\frac{1}{2} *\frac{1}{16}*16=\frac{1}{2}$

A number raised to a negative exponent is sometimes negative.

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great Valentines Day! :-)

- Cutiepatutie ☺❀❤

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A and E

Step-by-step explanation:

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What is an equation of the line that passes through the point (−1,−6) and is parallel to the line 5x-y=9

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Answer:

-5x + y = -1 or y = 5x - 1

Step-by-step explanation:

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7 0

3 years ago

Which pair of funtions is not a pair of inverse functions? please help!!

antiseptic1488 [7]

Answer:

$f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Step-by-step explanation:

we know that

To find the inverse of a function, exchange variables x for y and y for x. Then clear the y-variable to get the inverse function.

we will proceed to verify each case to determine the solution of the problem

case A) $f(x)=\frac{x+1}{6} , g(x)=6x-1$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y+1}{6}$

Isolate the variable y

$6x=y+1$

$y=6x-1$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=6x-1$

therefore

f(x) and g(x) are inverse functions

case B) $f(x)=\frac{x-4}{19} , g(x)=19x+4$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y-4}{19}$

Isolate the variable y

$19x=y-4$

$y=19x+4$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=19x+4$

therefore

f(x) and g(x) are inverse functions

case C) $f(x)=x^{5}, g(x)=\sqrt[5]{x}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=y^{5}$

Isolate the variable y

fifth root both members

$y=\sqrt[5]{x}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\sqrt[5]{x}$

therefore

f(x) and g(x) are inverse functions

case D) $f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y}{y+20}$

Isolate the variable y

$x(y+20)=y$

$xy+20x=y$

$y-xy=20x$

$y(1-x)=20x$

$y=20x/(1-x)$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=20x/(1-x)$

$\frac{20x}{1-x}\neq \frac{20x}{x-1}$

therefore

f(x) and g(x) is not a pair of inverse functions

7 0

3 years ago

Read 2 more answers

The mean for this data is 95 103,99,90,86,105,96

prisoha [69]

96.2857143 would be the answer because you add all the numbers together then divided by 7

5 0

3 years ago

Can someone tell me step by step how to solve this?

sasho [114]

Answer:

Step-by-step explanation:

I take it that this is some sort of ratio. Start by solving the brackets on the left.

Brackets: 1/3 - 1/10

Brackets: 10/30 - 3/30

Brackets: 7/30

Brackets^2: 49/900

Brackets^2 - 1/5: 49/900 - 180/900

Brackets^2 - 1/5: -131/900

(2/5)^2 = 4/25

The way this read, it should be

$\frac{\frac{-131}{900} }{\frac{4}{25} }$

Which when you invert and multiply becomes

$\frac{-131}{900} * \frac{25}{4}$

which finally becomes

$\frac{-131}{144}$

7 0

2 years ago