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3 years ago

Find the value of x.

Mathematics

2 answers:

Svetradugi [14.3K]3 years ago

8 0

For the value of x it would be 9

bija089 [108]3 years ago

3 0

Answer:

It's 3.

Step-by-step explanation:

If you divide 6 by 2, you would get 3.

Hope this helps!! (:

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Andru [333]

Answer:

$f^{-1}(x)=4x+20$

Step-by-step explanation:

Given

$f(x)=\frac{x}{4}-5\\ \\y=\frac{x}{4}-5$

Swap x and y and solve for y

$y=\frac{x}{4}-5\\ \\x=\frac{y}{4}-5\\ \\x+5=\frac{y}{4}\\ \\4(x+5)=y\\\\y=4x+20$

Final answer

$f^{-1}(x)=4x+20$

Keep in mind that we need to swap the variables because a function and its inverse are symmetrical about the line y=x. See the graph below.

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3 years ago

Which relation is a function?

Nadya [2.5K]

Answer:

it is c

Step-by-step explanation:

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3 years ago

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Dominik [7]

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7 0

3 years ago

I need to change the subject to x

tiny-mole [99]

(x - 4y) / w = 9

Multiply both sides by w

x - 4y = 9w

Add 4y to both sides

x = 9w + 4y

5 0

3 years ago

Tju [1.3M]

Answer:

Step-by-step explanation:

Rationalize the denominator of b. So, multiply the numerator and denominator by $\sqrt{x}$

$b = \frac{(1-2\sqrt{x}) *\sqrt{x}}{\sqrt{x}*\sqrt{x} }=\frac{1*\sqrt{x} -2\sqrt{x} *\sqrt{x} }{\sqrt{x} *\sqrt{x} }\\\\=\frac{\sqrt{x} -2x}{x}\\$

Now, find a +b

$a +b = \frac{2x+\sqrt{x} }{x}+\frac{\sqrt{x} -2x}{x}\\\\=\frac{2x+\sqrt{x} +\sqrt{x} -2x}{x}$

Combine like terms

$= \frac{2x-2x+\sqrt{x} +\sqrt{x} }{x}\\\\=\frac{2\sqrt{x} }{x}$

Now find (a + b)²

(a +b)² = $(\frac{2\sqrt{x} }{x})^{2}$

$= \frac{2^{2}*(\sqrt{x} )^{2}}{x^{2}}\\\\= \frac{4* x}{x^{2}}\\\\= \frac{4}{x}$

Hint: $\sqrt{x} *\sqrt{x} =\sqrt{x*x}=x$

5 0

3 years ago