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insens350 [35]
3 years ago
8

Two cars are 190 miles apart and travel toward each other along the same road. They meet in 2 hours. One car travels 5 miles per

hour faster than the other car. What is the average speed of each car?
Mathematics
1 answer:
horrorfan [7]3 years ago
6 0

The speed of slower car is 45 mph and faster car is 50 mph.

Step-by-step explanation:

Given,

Distance covered by both cars = 190 miles

Time taken by both of them = 2 hours

Let,

x be the speed of slower car.

Speed of faster car = x+5

Combined speed = x+x+5 = 2x+5

We know that;

Distance = Speed * Time

190=(2x+5)*2\\190=4x+10\\4x=190-10\\4x=180\\

Dividing both sides by 4

\frac{4x}{4}=\frac{180}{4}\\x=45

Speed of slower car = 45 miles per hour

Speed of faster car = 45+5 = 50 miles per hour

The speed of slower car is 45 mph and faster car is 50 mph.

Keywords: speed, distance

Learn more about speed at:

  • brainly.com/question/909731
  • brainly.com/question/902892

#LearnwithBrainly

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vazorg [7]
The idea being, when you run a perpendicular line to the base, from the right-angle in a right-triangle, like in this case, what you end up with is, three similar triangles, a Large triangle containing the two smaller ones, a Medium triangle and a Small one, check the picture below.

since all three are similar, we can use the proportions on the corresponding sides, as you see in the picture, the base of the triangle then will just be x + y.

5 0
3 years ago
Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro
Ugo [173]

Answer:

Side\ B = 6.0

\alpha = 56.3

\theta = 93.7

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with \alpha, \beta, \theta

[See Attachment for Triangle]

A = 10cm

C = 12cm

\beta = 30

What the question is to calculate the third length (Side B) and the other 2 angles (\alpha\ and\ \theta)

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

B^2 = A^2 + C^2 - 2ABCos\beta

Substitute: A = 10,  C =12; \beta = 30

B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30

B^2 = 100 + 144 - 240*0.86602540378

B^2 = 100 + 144 - 207.846096907

B^2 = 36.153903093

Take Square root of both sides

\sqrt{B^2} = \sqrt{36.153903093}

B = \sqrt{36.153903093}

B = 6.0128115797

B = 6.0 <em>(Approximated)</em>

Calculating Angle \alpha

A^2 = B^2 + C^2 - 2BCCos\alpha

Substitute: A = 10,  C =12; B = 6

10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha

100 = 36 + 144 - 144 *Cos\alpha

100 = 36 + 144 - 144 *Cos\alpha

100 = 180 - 144 *Cos\alpha

Subtract 180 from both sides

100 - 180 = 180 - 180 - 144 *Cos\alpha

-80 = - 144 *Cos\alpha

Divide both sides by -144

\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}

\frac{-80}{-144} = Cos\alpha

0.5555556 = Cos\alpha

Take arccos of both sides

Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)

Cos^{-1}(0.5555556) = \alpha

56.25098078 = \alpha

\alpha = 56.3 <em>(Approximated)</em>

Calculating \theta

Sum of angles in a triangle = 180

Hence;

\alpha + \beta + \theta = 180

30 + 56.3 + \theta = 180

86.3 + \theta = 180

Make \theta the subject of formula

\theta = 180 - 86.3

\theta = 93.7

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3 years ago
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Aleks [24]
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- (2 T +   P = 40 )
________________

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P = 80/4
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Agata [3.3K]

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3 years ago
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