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sweet-ann [11.9K]
2 years ago
10

The train traveled at a speed of 90 miles per hour for 3 hours then slowed down to 72 miles per hour for the next 1 1/2 hours. W

hat is the total number of miles the train traveled?
Mathematics
1 answer:
Viefleur [7K]2 years ago
4 0

Answer:

378 miles

Step-by-step explanation:

<u>Remember that (rate)(distance) = time</u>

The train is going 90 mph for 3 hours. That's 270 miles in 3 hours.

Then you would multiply 72 (mph) by 1 1/2 (hours) and get 108 miles traveled in 1 1/2 hours.

Add those together and you get 378.

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<em>Answer</em><em> </em><em>is option</em><em> </em><em>c</em>

Step-by-step explanation:

given \: equations \: are \\ x - 2y = 15 \:  \: and \: 2x + 4y =  - 18 \\ divide \: by \: 2 \: in \: second \: equation \\ x  + 2y =  - 9 \:  \: (equation \: 3) \\ adding \: 1and \: 3 \: we \: get \\ 2x = 6 \\ x = 3 \\ substitute \: x = 3 \: in \: equation \: 3 \\ 3 + 2y =  - 9 \\ 2y =  - 9 - 3 \\ 2y =  - 12 \\ y =  - 6

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olga55 [171]
We know that:

(x-a)^2+(y-b)^2=r^2

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When we substitute x and y (from the pairs we have), we'll get a system of equations:

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There will be:

\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\&#10;\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\&#10;\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\&#10;

From equations (II) and (III) we have:

\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}

and from (I) and (II):

\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}

Now we can easly calculate a and b:

\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}

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(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}
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