All categories

3 years ago

Can someone help me with this question please. I've been struggling.

Mathematics

1 answer:

Thepotemich [5.8K]3 years ago

7 0

Answer:

y=-1

Step-by-step explanation:

Since we see an overlap, the line must cross through the original image.
We can solve the equation of the line of reflection by realizing there is no slope because the reflection is parallel to the x-axis.

You might be interested in

PLEASE HELP ME I NEED THIS ANSWER FOR ALGEBRA!

Marizza181 [45]

Answer:

$radius = 6.2$

Step-by-step explanation:

so you already have the formula, which is

$\sqrt[3]{ \frac{3v}{4\pi} }$

the v represents Volume.

and 3v would be 3×volume.

$\sqrt[3]{ \frac{3 \times 1000}{4 \times \pi} }$

$\sqrt[3]{ \frac{3000}{12.56637061} }$

$\sqrt[3]{238.7324147 }$

$= 6.20350491$

to the nearest tenth place.

$= 6.2$

4 0

3 years ago

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

L = luminosity of the star (related to the Sun)

d = distance in ly (light-years)

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$