If point S is reflected across the y axis what are thecoordinates of its reflection image

1 answer:

Ivan3 years ago

3 0

What are the coordinates of point S? Whatever they are the answer should be (-x, y) depending on the location of your point.

You might be interested in

Frank is going to plant y vegetable seeds in one garden and 2y + 10 vegetable seeds in another.

Sholpan [36]

Answer:

$3y + 10$

Step-by-step explanation:

The total number of seeds Frank will plant in both gardens all together would be:

$y + 2y + 10$

Add like terms together to simplify the expression

$3y + 10$

Frank is going to plant $3y + 10$ seeds

3 0

3 years ago

Find the value of x and y

KiRa [710]

Answer:

y=57 degrees and x=66 degrees

Step-by-step explanation:

123+57 =180 so angle y=57 degrees

57+57=114 all triangles = 180 degrees so 180-114=66 degrees=X

4 0

3 years ago

Derivative of these questions<br>

shtirl [24]

Answer: $\bold{y'=\dfrac{1}{2(a-b)^2}\bigg(\dfrac{1}{\sqrt{x+a}}+\dfrac{1}{\sqrt{x+b}}\bigg)}$

<u>Step-by-step explanation:</u>

$y=\dfrac{1}{\sqrt{x+a}-\sqrt{x+b}}\\\\\\\text{Rationalize the denominator:}\\y=\dfrac{1}{\sqrt{x+a}-\sqrt{x+b}}\bigg(\dfrac{\sqrt{x+a}+\sqrt{x+b}}{\sqrt{x+a}+\sqrt{x+b}}\bigg)\\\\\\y=\dfrac{\sqrt{x+a}+\sqrt{x+b}}{x+a-x-b}\\\\\\y=\dfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}\\\\\\\text{Apply the derivative (derivative of the numerator (top) divided by the}\\\text{denominator (bottom) squared).}$

$\text{Derivative of }\sqrt{x+a}\quad \rightarrow \quad (x+a)^{\frac{1}{2}}\quad = \quad \dfrac{1}{2}(x + a)^{-\frac{1}{2}}\\\\\text{Derivative of }\sqrt{x+b}\quad \rightarrow \quad (x+b)^{\frac{1}{2}}\quad = \quad \dfrac{1}{2}(x + b)^{-\frac{1}{2}}\\\\\\\text{Derivative of}\ \dfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}:\\\\= \dfrac{\dfrac{1}{2}(x + a)^{-\frac{1}{2}}+\dfrac{1}{2}(x + b)^{-\frac{1}{2}}}{(a-b)^2}$

$\text{Factor out }\dfrac{1}{2}\ \text{from the numerator:}\\\\\dfrac{(x+a)^{-\frac{1}{2}}+(x+b)^{-\frac{1}{2}}}{2(a-b)^2}\\\\\\\text{Move the terms with negative exponents to the denominator:}\\\\=\dfrac{1}{2(a-b)^2}\bigg(\dfrac{1}{\sqrt{x+a}}+\dfrac{1}{\sqrt{x+b}}\bigg)$