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3 years ago

Chelsea wants to buy a new iPhone that costs $324. She

Mathematics

2 answers:

vampirchik [111]3 years ago

5 0

Answer:

13.4 or about two weeks

Step-by-step explanation:

so if she already has $56 and she needs $268 more to buy the phone so we'll do 268 divided by 20 to see how many weeks it will take which is 13.4 so you can either put 13.4 as the answer or you can put about two weeks because there are 7 days in a week and 13.4/7 is ~1.9 or 2

bulgar [2K]3 years ago

4 0

Answer:

14 full weeks.

Step-by-step explanation:

The phone costs $324 and she saved $56, so we subtract $56 from $324.

$324 - $56 = $268

She saves $20 each week so we divide $268 by $20 to get the number of weeks.

$268/20 = 13.4

It will take 13.4 weeks, but we have to round up because at 13 weeks she won't have enough saved. So 14 full weeks.

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krek1111 [17]

8*4=32
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3 years ago

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1.) -49

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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

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

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

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