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Ainat [17]
3 years ago
9

Two cars leave Phoenix and travel along roads 90 degrees apart. If Car 1 leaves 30 minutes earlier than Car 2 and averages 42 mp

h and if Car 2 averages 50 mph, how far apart will they be after Car 1 has traveled 3.5 hours?
A: 230
B: 250
C: 210
Mathematics
1 answer:
romanna [79]3 years ago
3 0

Answer:

C. 210 miles

Step-by-step explanation:

We have been given that two cars leave Phoenix and travel along roads 90 degrees apart. Car 1 leaves 30 minutes earlier than Car 2 and averages 42 mph.

We will use distance formula and Pythagoras theorem to solve our given problem.

\text{Distance}=\text{Speed}\times \text{Time}

\text{Distance covered by Car 1 in 3.5 hours}=\frac{42\text{ Miles}}{\text{Hour}}\times \text{3.5 hour}

\text{Distance covered by Car 1 in 3.5 hours}=147\text{ Miles}.

Since car 1 leaves 30 minutes before car 2, so car 2 will travel for only 3 hours when car 1 will travel for 3.5 hours.

\text{Distance covered by Car 2 in 3 hours}=\frac{50\text{ Miles}}{\text{Hour}}\times \text{3 hour}

\text{Distance covered by Car 2 in 3 hours}=150\text{ Miles}

Since both car travel along roads 90 degree apart, therefore, the distance between both cars after Car 1 has traveled 3.5 hours would be hypotenuse with legs 147 and 150.

\text{Distance between both cars}=\sqrt{147^2+150^2}

\text{Distance between both cars}=\sqrt{21609+22500}

\text{Distance between both cars}=\sqrt{44109}

\text{Distance between both cars}=210.021427\approx 210

Therefore, the both cars will be 210 miles apart and option C is the correct choice.

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Answer:

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Step-by-step explanation:

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How long does Trevon needs to drive before they are 65km apart.

Note : both took at the same time and in opposite direction.

Recall : speed = distance / time

Distance = speed * time

Distance between them = 65km

Let Trevon's distance = D2 = 43 * t

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4 years ago
For the equation given below, evaluate y′ at the point (−2,2)<br> xe^y−4y=2x−4−2e^2
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Implicit differentiation
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I'll show the steps partially
e^y+xe^yy'-4y'=2
xe^yy'-4y'=2-e^y
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Which is not an equation of the line going through (6,7) and (2,-1) A. y-7=2(x-6) B. y=2x-5 C. y-1=2(x+2) D. y+1=2(x-2)
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Answer:

C. y-1=2(x+2)

Step-by-step explanation:

A. y-7=2(x-6)-point slope form: y-y1=2(x-x1)-coordinate point: (x1,y1)=(6,7)

Now let's check that the other point works-just sub in the coordinates for the x and y variables.

y-7=2(x-6)\\-1-7=2(2-6)\\-8=2(-4)\\-8=-8

So answer choice A. goes through both points, so no.

B. y=2x-5-slope-intercept form, so just sub in both sets of coordinates separately to see if they each work.

y=2x-5\\7=2(6)-5\\7=12-5\\7=7\\equal

y=2x-5\\-1=2(2)-5\\-1=4-5\\-1=-1\\equal

Both work, so no.

C. y-1=2(x+2)-point slope form that gives us first coordinate set--(-2,1)

Except that the point is actually (2,-1) not (-2,1).  That means that answer choice C is not an equation of the line thaat goes through (6,7) and (2,-1), so yes.

D. Anyway, if we check D--y+1=2(x-2)--coordinate point given: (2,-1), which is correct, so no.

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