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3 years ago

I will give u brainliest

Mathematics

1 answer:

wolverine [178]3 years ago

5 0

Answer:

≈ 7.35

The square root of 54 is close to 7.

7 squared = 49 and that is close to 54, but the square root of 54 is slightly larger

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2 years ago

What is the 100 nearest 513

ololo11 [35]

Answer: 500

Step-by-step explanation:

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3 years ago

Find x, if 3x − 5(y + x ) ^2 = 16 and 4x + 2(y + x )^ 2 = 30

lana66690 [7]

Answer:

x = 7

Step-by-step explanation:

Start by solving for $(y+x)^2$ which is a common expression for both equations. Then in the first equation:

$3\,x5\,(y+x)^2=16\\3x-16=5\,(y+x)^2\\(y+x)^2=\frac{3\,x-16}{5}$

and in the second equation:

$4\,x+2\,(y+x)^2=30\\2\,(y+x)^2=30-4\,x\\(y+x)^2=15-2\,x$

Now we make the two $(y+x)^2$ expressions equal so we get:

$\frac{3\,x-16}{5} =15-2\,x\\3\,x-16=75-10\,x\\13\,x=75+16\\13\,x=91\\x=7$

8 0

3 years ago

2x – y = 6 5x + 10y = –10

Studentka2010 [4]

2x-y=6;5x+10y=-10. -y=-2x+6. -y/-1=-2x+6/-1. y=2x-6. 5x+10y=-10. 5x+10(2x-6)=-10. 25x-60=-10. 25x=50. 25x/25=50/25. y=2x-6. y=(2)(2)-6. y=-2. Therefore, x=2 and y=-2

8 0

2 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

<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

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

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0

3 years ago