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Zinaida [17]
3 years ago
15

A map has a scale of 1 cm : 11.5 km. Two cities are 5.5 cm apart on the map. To the nearest tenth of a kilometer, what is the ac

tual distance corresponding to the map distance?
A. 74.8 km

B. 120.8 km

C. 68.8 km

D. 63.3 km
Mathematics
2 answers:
galina1969 [7]3 years ago
7 0

Answer:

the answer is 63.3 km


777dan777 [17]3 years ago
5 0

All that you need to do here is to multiply 5.5 by 11.5 which equates to...

=5.5x11.5

=63.25

because the second decimal is five you have to round up so that means that the answer is 63.3km

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Answer:

x1= -6 x2=6

Step-by-step explanation:

x^2-10 = 26

x^2 = 26 +10

x^2 = 36

x = sqrt(36)

x = -6 or x = 6 because both 6 and -6 squared are equal to 36

3 0
3 years ago
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What is the solution to 5(3-x)?
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The answer is 15 - 5 x

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I hope this helps and please mark brainliest!
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2 years ago
Will Mark Brainlest. (find the domain of following relation for the given range) step by step ​
mr_godi [17]

Step-by-step explanation:

I suppose you mean find the range for the following domain. Because when

g(x) is given the value of X

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If you want to solve for the domain, the question should give

g(x) = y

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X³+X²-X-2

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Substitute

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8+4-2-2

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5 0
2 years ago
Solve on [0,2pi). Answer in order from smallest to biggest<br> cos x sin x = sin x
Musya8 [376]

Answer:

0, π

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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.

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3 years ago
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