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3 years ago

Please help, i do not get this.

Mathematics

1 answer:

Artyom0805 [142]3 years ago

3 0

Answer:

k(x) = |x - 4| - 2

Step-by-step explanation:

So we know that k(x) is shifted 4 untis to the right and 2 units down. In an absolute value function, the value you move to the left will be positive and the value you move to the right will be negative. Since we are moving to the right, our value will be negative.

When you shift the function up or down, you will either subtract or add that value out side the absolute value. Since we are shifting down 2 units, that means we will be subtracting by 2.

Best of Luck!

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How much do you save per class by taking the longer-term class: Weight Training, $200 for 18 weeks (54 classes) or $350 for 36 w

Vsevolod [243]

Shorter term class:

You pay $200 for 54 classes in total. So,

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5 0

3 years ago

Read 2 more answers

What is the equation of the line perpendicular to 3x+y= -8that passes through -3,1? Write your answer in slope-intercept form. S

Gekata [30.6K]

Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is $y=\frac{1}{3} x+2$

Solution:

Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).

Generic slope intercept form of a line is given by y = mx + c

where m = slope of the line.

Let's first find slope intercept form of 3x + y = -8

3x + y = -8

=> y = -3x - 8

On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3

And as the line passing through (-3,1) and is perpendicular to 3x + y = -8, product of slopes of two line will be -1 as lies are perpendicular.

Let required slope = x

$\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}$

So we need to find the equation of a line whose slope is $\frac{1}{3}$ and passing through (-3,1)

Equation of line passing through $(x_1 , y_1)$ and having lope of m is given by

$\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)$

$\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}$

Substituting the values we get,

$\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}$