All categories

3 years ago

Graph Jkl and its image after a reflection in the given line J(5,3), K(1,-2), l(-3,4); y-axis

Mathematics

1 answer:

Sergeeva-Olga [200]3 years ago

7 0

The reflection (x,y) over the y-axis is (-x,y)
So this yields J(-5,3), K(-1,-2), I(3,4)

You might be interested in

Which describes the slope of this line?

Mice21 [21]

Positive Slope

TIP for finding what slope it is:

Positive - the line is pointing right

Negative - the line is pointing left

Undefined - the line is pointing up and down

Zero - the line is point from left to right

4 0

3 years ago

Read 2 more answers

How many centimeters are in 5 inches? 1in.=2.54cm

Tanya [424]

Answer:

12.7 inches

Step-by-step explanation:

you need to multiply 2.54*5

and your final answer is 12.7 in

8 0

3 years ago

How do I differentiate this?

disa [49]

Answer:

$\frac{8}{\left(-x+2\right)^2}$ & $x = 1, x = 3$

Step-by-step explanation:

I'm sure you are familiar with the product rule,

If y = u*v => dy/dx = u * dv/dx + v * dy/dx <----- product rule

In this case:

$u=3x+2,\:v=\left(2-x\right)^{-1},\\=>\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)$

Now remember the sum rule:

$\frac{d}{dx}\left(3x+2\right) = \frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(2\right),\\\frac{d}{dx}\left(3x\right) = 3,\\\frac{d}{dx}\left(2\right) = 0\\\frac{d}{dx}\left(3x+2\right) = 3$

For this second bit we apply the chain rule:

$\frac{d}{dx}\left(\left(2-x\right)^{-1}\right) = -\frac{1}{\left(2-x\right)^2}\frac{d}{dx}\left(2-x\right),\\\frac{d}{dx}\left(2-x\right) = -1,\\\\=> -\frac{1}{\left(2-x\right)^2}\left(-1\right)\\=> \frac{1}{\left(2-x\right)^2}$

If we substitute these values back into the expression... $\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)$

...we get the following:

$3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)$

The rest is just pure simplification:

$3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)\\= \frac{3}{-x+2}+\frac{3x+2}{\left(-x+2\right)^2}\\= \frac{3\left(-x+2\right)}{\left(-x+2\right)^2}+\frac{3x+2}{\left(-x+2\right)^2}\\\\= \frac{3\left(-x+2\right)+3x+2}{\left(-x+2\right)^2}\\\\= \frac{8}{\left(-x+2\right)^2}$

Now let's equate this to equal 8 for the second bit and solve for x:

$\frac{8}{\left(-x+2\right)^2}=8,\\\frac{8}{\left(-x+2\right)^2}\left(-x+2\right)^2=8\left(-x+2\right)^2,\\8=8\left(-x+2\right)^2,\\\left(-x+2\right)^2=1,\\x = 1, x = 3$