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OLEGan [10]
3 years ago
12

2 cars raced at a track. the faster car traveled 20mph faster than the slower car. in the time that the slower car traveled 165

miles, the faster car traveled 225 miles. if the speeds of the cars remained constant, how fast did the slower car travel during the race.
Mathematics
1 answer:
icang [17]3 years ago
3 0

Answer:

The speed of slower car is 55 miles per hour.

Step-by-step explanation:

Given as :

The speed of slower car = s_2 = s  mph

The speed of faster car = s_1 = ( s + 20 ) mph

The distance cover by slower car = d_2 = 165 miles

The distance cover by faster car = d_1 = 225 miles

The time taken by both cars for travelling = t hours

The speed of the cars remains constant

Now, <u>According to question</u>

∵ Time = \dfrac{\textrm Distance}{\textrm Speed}

So,<u> For slower car</u>

t = \dfrac{d_2}{s_2}

Or, t = \dfrac{165}{s}           ............1

So, For faster car

t = \dfrac{d_1}{s_1}

Or, t = \dfrac{225}{s+20}           ............2

Now, equating both the equations

I.e  \dfrac{225}{s+20} = \dfrac{165}{s}  

<u>By cross multiplying</u>

Or, 225 × s = 165 × (s + 20)

Or, 225 s = 165 s + 3300

Or, 225 s - 165 s = 3300

Or, 60 s = 3300

∴  s = \dfrac{3300}{60}

I.e s = 55 miles per hour

So , The speed of slower car = s_2 = s = 55 miles per hour

Hence , The speed of slower car is 55 miles per hour.  Answer

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Answer:

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Step-by-step explanation:

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multiply both sides by "2 bags" to isolate the unknown X

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8 0
2 years ago
12a + 12b + 12c + 12d Michael writes the expression shown here as 12(a + b + c + d). He is correctly using the ___________ prope
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Answer:

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Step-by-step explanation:


3 0
3 years ago
8.Find the value of c that makes the expression be a perfect square trinomial p^2-30p+c
Darina [25.2K]
Hope this is the right one.

Problem 8
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<em>Step One</em>
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1/2 * -30 = - 15

<em>Step 2
</em>Square -15
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Problem Nine
\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a} &#10;
a = 1
b = 4
c = -15
\text{x = }\dfrac{ -4 \pm \sqrt{4^{2} - 4*1*(-15) } }{2*1}
\text{x = }\dfrac{ -4 \pm \sqrt{\text{16} + \text{60} } }{2}

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x = [-4 +/- 2*sqrt19]/2
x = [-4/2 +/- 2/2 sqrt[19]
x = - 2 +/- sqrt(19) 

x1 = - 2 + sqrt(19)
x2 = -2 - sqrt(19)

These two can be broken down more by finding the square root. I will leave them the way they are.  It's just a calculator question if you want it to go into decimal form.

Problem Ten

a = 1
b = 4
c = -32

The discriminate is sqrt(b^2 - 4ac)
D = sqrt(b^2 - 4ac)
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D = +/- 12

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(x + 8)(x - 4) = y 
But that' s not what you were asked for.
The discriminate is >  0 so the roots are going to be real.
<em>Answer; The discriminate is > 0 so there will be 2 real different roots.</em>

4 0
3 years ago
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He still had 600 left to go...

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He had to run three times longer than Eric...

3000/3=1000

Does that make sense?
8 0
3 years ago
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vredina [299]

9514 1404 393

Answer:

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Step-by-step explanation:

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6 0
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