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3 years ago

PLEASE HELP WILL MARK BRAINLIEST!!

Mathematics

2 answers:

Komok [63]3 years ago

8 0

Answer:

Hello!

1st problem: x = 144

2nd problem: x = 14

3rd problem: x = 144

4th problem: x = 14

Gre4nikov [31]3 years ago

7 0

Answer:

Explanation below.

Step-by-step explanation:

1. a. $\sqrt{x} =12$

b. $\sqrt{x} ^{2}=12^{2}$

c. x=144

2. a. $\sqrt{x^{2} } =\sqrt{196}$

b. x= 14

$\sqrt{x} -4 +4= 8+4\\\sqrt{x} =12\\x= 3.46410161514 \\or \\2\sqrt{3}$

$2(x^{2} +3)=398\\x^{2} +3= 199\\x^{2} =196\\x=14$

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If six times a number is decreased by fourteen, the result is 124. What is the number

denis23 [38]

Answer:

$\large\boxed{23}$

Step-by-step explanation:

Six times a number is decreased by fourteen, the result is 124.

Write an equation that represents this

(6 * x) - 14 = 124

Multiply 6 by x to remove the parenthesis

6x - 14 = 124

Add 14 to both sides of the equation

6x = 138

Divide both sides of the equation by 6

x = 23

The unknown number is $\large\boxed{23}$

Hope this helps :)

3 0

3 years ago

Read 2 more answers

If you answer all the questions you get brainliest

Triss [41]

The ans is the one with G'(-6,0)

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3 years ago

Which statement describes the inverse of m(x) = x^2 – 17x?

DochEvi [55]

Given:

The function is

$m(x)=x^2-17x$

To find:

The inverse of the given function.

Solution:

We have,

$m(x)=x^2-17x$

Substitute m(x)=y.

$y=x^2-17x$

Interchange x and y.

$x=y^2-17y$

Add square of half of coefficient of y , i.e., $\left(\dfrac{-17}{2}\right)^2$ on both sides,

$x+\left(\dfrac{-17}{2}\right)^2=y^2-17y+\left(\dfrac{-17}{2}\right)^2$

$x+\left(\dfrac{17}{2}\right)^2=y^2-17y+\left(\dfrac{17}{2}\right)^2$

$x+\left(\dfrac{17}{2}\right)^2=\left(y-\dfrac{17}{2}\right)^2$ $[\because (a-b)^2=a^2-2ab+b^2]$

Taking square root on both sides.

$\sqrt{x+\left(\dfrac{17}{2}\right)^2}=y-\dfrac{17}{2}$

Add $\dfrac{17}{2}$ on both sides.

$\sqrt{x+\left(\dfrac{17}{2}\right)^2}+\dfrac{17}{2}=y$

Substitute $y=m^{-1}(x)$ .

$m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}$

We know that, negative term inside the root is not real number. So,

$x+\left(\dfrac{17}{2}\right)^2\geq 0$

$x\geq -\left(\dfrac{17}{2}\right)^2$

Therefore, the restricted domain is $x\geq -\left(\dfrac{17}{2}\right)^2$ and the inverse function is $m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}$ .

Hence, option D is correct.

Note: In all the options square of $\dfrac{17}{2}$ is missing in restricted domain.

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brilliants [131]

Answer:

Step-by-step explanation:

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