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2 years ago

Given h(x) = -X – 2, solve for x when h(x) = 3.

Mathematics

2 answers:

viktelen [127]2 years ago

8 0

Answer:

x = -5

Step-by-step explanation:

One is given a function and the value of the function at a certain output. One must set the function to the given output, and solve to find the input,

h(x) = - x - 2

h(x) = 3

Set the output equal to the input,

h(x) = - x - 2 = 3

Inverse operations,

- x - 2 = 3

+2 +2

- x = 5

*-1 *-1

x = -5

Naya [18.7K]2 years ago

6 0

The answer should be h(3)= -5

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What is the result of a dilation of scale factor 3 centered at the origin of the line 2y + 3x=10?? PLEASE HELP PLEASEEEEEEEEE

maks197457 [2]

Given:

The equation of a line is:

$2y+3x=10$

The line is dilated by factor 3.

To find:

The result of dilation.

Solution:

The equation of a line is:

$2y+3x=10$

For $x=0$ ,

$2y+3(0)=10$

$2y+0=10$

$y=\dfrac{10}{2}$

$y=5$

For $x=2$ ,

$2y+3(2)=10$

$2y+6=10$

$2y=10-6$

$2y=4$

Divide both sides by 2.

$y=\dfrac{4}{2}$

$y=2$

The given line passes through the two points A(0,5) and B(2,2).

If the line dilated by factor 3 with origin as center of dilation, then

$(x,y)\to (3x,3y)$

Using this rule, we get

$A(0,5)\to A'(3(0),3(5))$

$A(0,5)\to A'(0,15)$

Similarly,

$B(2,2)\to B'(3(2),3(2))$

$B(2,2)\to B'(6,6)$

The dilated line passes through the points A'(0,15) and B'(6,6). So, the equation of dilated line is:

$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

$y-15=\dfrac{6-15}{6-0}(x-0)$

$y-15=\dfrac{-9}{6}(x)$

$y-15=\dfrac{-3}{2}x$

Multiply both sides by 2.

$2(y-15)=-3x$

$2y-30=-3x$

$2y+3x=30$

Therefore, the equation of the line after the dilation is $2y+3x=30$ .

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