Ask question

Ask question

All categories

2 years ago

13

If the graph of Rx) = |x| is shifted up 7 units, which equation represents the new graph? OA. g(x) = (x + 7|| O B. g(x) = |x-71

O C. g(x) = |x] -7 OD. g(x) = |x| + 7

1 answer:

Tju [1.3M]2 years ago

8 0

Answer:

answer B

Step-by-step explanation:

hope this helps

You might be interested in

Explain how you change a decimal to a fraction.

maw [93]

Answer:

Decimals can be written in fraction form. To convert a decimal to a fraction, place the decimal number over its place value. For example, in 0.6, the six is in the tenths place, so we place 6 over 10 to create the equivalent fraction, 6/10. If needed, simplify the fraction.

7 0

3 years ago

Read 2 more answers

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

Segment FG begins at point F(-2, 4) and ends at point G (-2, -3). Segment FG is translated by (x, y) → (x – 3, y + 2) and then r

Mariana [72]

Answer:

The length of the segment F'G' is 7.

Step-by-step explanation:

From Linear Algebra we define reflection across the y-axis as follows:

$(x',y')=(-x, y)$ , $\forall\, x, y\in \mathbb{R}$ (Eq. 1)

In addition, we get this translation formula from the statement of the problem:

$(x',y') =(x-3,y+2)$ , $\forall \,x,y\in \mathbb{R}$ (Eq. 2)

Where:

$(x, y)$ - Original point, dimensionless.

$(x', y')$ - Transformed point, dimensionless.

If we know that $F(x,y) = (-2, 4)$ and $G(x,y) = (-2,-3)$ , then we proceed to make all needed operations:

Translation

$F''(x,y) = (-2-3,4+2)$

$F''(x,y) = (-5,6)$

$G''(x,y) = (-2-3,-3+2)$

$G''(x,y) = (-5,-1)$

Reflection

$F'(x,y) = (5, 6)$

$G'(x,y) = (5,-1)$

Lastly, we calculate the length of the segment F'G' by Pythagorean Theorem:

$F'G' = \sqrt{(5-5)^{2}+[(-1)-6]^{2}}$

$F'G' = 7$

The length of the segment F'G' is 7.

8 0

2 years ago

3/4, 4/5, 7/8, 9/11, 5/6<br>can they be simplified? give your reasons below.

Bad White [126]

They cannot be simplified

4 0

2 years ago

Read 2 more answers

A coupon subtracts $17.95 from the price p of a pair of headphones. You pay $71.80 for the headphones after using the coupon. Wr

Volgvan

$264.75 I like math it took 15 minutes but I got it

7 0

3 years ago

Read 2 more answers