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3 years ago

8

Ill love u forever if u answer correct Point F is reflected over the y-axis to create F’. Use an ordered pair to name the locati

on of F’, and determine the distance between F and F’.

1 answer:

andrew-mc [135]3 years ago

4 0

Answer:

F' = (-6,3)

Distance = 12 units

Step-by-step explanation:

F is currently at (6,3).

Reflected over the y-axis, it would be at (-6,3).

That puts them 12 units apart.

You might be interested in

A watermelon seed is 0.4cm long. How far would a line of 126 stretch?

nikdorinn [45]

Answer:

0.504m

Step-by-step explanation:

First, we would multiply 0.4cm by 126 watermelon seeds to find the length of the line in cm.

0.4*126

= 50.4cm

Therefore, the line of watermelon seeds stretches 50.4cm.

Now, because the question wants us to type the answer in meters, we must convert 50.4cm into meters. We do this by dividing 50.4 by 100 since 100cm makes up 1 meter.

50.4/100

= 0.504m

Therefore, the line of watermelon seeds stretches 0.504m.

I hope this helps!

6 0

3 years ago

Compare the surface area-to-volume ratios of the moon and mars. express your answer using two significant figures.

antoniya [11.8K]

we know that

For a spherical planet of radius r, the volume V and the surface area SA is equal to

$V=\frac{4}{3} *\pi *r^{3} \\ \\ SA=4*\pi *r^{2}$

The
ratio of these two quantities may be written as

$SAV =\frac{(4*\pi*r^{2})}{(\frac{4}{3}*\pi*r^{3})} \\ \\ SAV =\frac{3}{r}$

we know

$rMoon=1,738Km\\ rMars=3,397 Km$

$\frac{SAV Moon}{SAV Mars} =\frac{3}{rMoon} *\frac{rMars}{3} \\ \\ \frac{SAV Moon}{SAV Mars} =\frac{rMars}{rMoon} \\ \\ \frac{SAV Moon}{SAV Mars} =\frac{3,397}{1,738} \\ \\ \frac{SAV Moon}{SAV Mars} =1.9545$

therefore

the answer is

$1.9$

5 0

3 years ago

Read 2 more answers

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

3 years ago

Select the correct answer.

mamaluj [8]

Answer:

C.$0.60

hope this help!!!!!!!!!!!!!!!!!!!!!!

3 0

2 years ago

Help needed with these for problems

Tatiana [17]

I can't read the first one and I also don't know

4 0

3 years ago