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3 years ago

13

Please help I don't know how to explain

1 answer:

spin [16.1K]3 years ago

7 0

Well, 1/4 = 2x.
1/4 / x = 2

cause 2x is multiplication
and the other side shows divison for the right side of the and

also, solution
x = .125

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Between 4,632 and 20,000 is what number?*

yarga [219]

To find the mid-way number we do this:

find the difference between the highest number and the lowest given number.
Divide the difference by 2
Add that number to the lowest number or subtract that number from the highest number.

20,000 - 4,632 = 15,362
$\frac{15,368}{2}$ = 7,684

4,632 + 7,684 = 12,316
or (we get the same number)
20,000 - 7,684 = 12, 316

12,316 is between 4,632 and 20,000

7 0

3 years ago

Can someone please help me with this

velikii [3]

Answer:

The answer is D.) Quadrant IV

Step-by-step explanation:

I took a quiz on this before and Quadrant IV was correct but I don't know if it's true because some of you're answers are different from mine but on my quiz D was the answer

8 0

2 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

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

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

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

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

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

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

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

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

I need help with this<br>histogram

chubhunter [2.5K]

A is what I thinkkkk

4 0

3 years ago

What is the initial value of the function represented by this table?

lakkis [162]

Answer:

Option D. 5

Step-by-step explanation:

we know that

The initial value of the function is the y-intercept, so is the value of the function when the value of x is equal to zero

In this problem observing the table

For x=0

The value of y is equal to y=5

therefore

The initial value is 5

4 0

3 years ago