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2 years ago

5

The base of the isosceles triangle is parallel to x-axis and has both end-points on the parabola y=x(10−x) and its vertex belong

s to the x-axis. Find area of the triangle if the length of its base is 8.

1 answer:

baherus [9]2 years ago

6 0

Answer:

36

Step-by-step explanation:

We must determine the x and y intercepts of the parabola:

When y=0, x=0 or x=10

WE know that the point of the triangle base is x and x+8. We can substitute this into the parabola equation because the endpoints are on the parabola.

f(x+8) $=-(x^2+16x+64)+10x+80$

f(x+8) $=-x^2-6x+16$

f(x)=f(x+8)

$-x^2+10x=-x^2-6x+16$

solve for x

$16x=16$

$x=1$

Therefore the heigh is f(1):

$=-1^2+10=9$

The area of the triangle is 1/2 base x height:

$=(1/2)\cdot{8}\cdot{9}=36$

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Answer:

The value is $k =109.6 \ miles$

Step-by-step explanation:

The diagram illustrating the question is shown on the first uploaded image

From the question we are told that

The distance from city A to B is AB = 467.3 miles

The bearing from B to C is $\theta_{BC} = N 28.7E$

The bearing from B to A is $\theta_{BA} = N 60.7E$

The bearing from A to B is $\theta_{AB} = S60.7W$

The bearing from A to C is $\theta_{AC} = S79.1W$

Generally from the diagram

$\theta_A = 180 - 60.7 -79.1$

=> $\theta_A = 40.2 ^o$

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=> $\theta_C = 180 - (40.2 + 32 )$

=> $\theta_C = 107.8 ^o$

Generally according to Sine Rule

$\frac{BC}{sin (\theta_A)} = \frac{CA}{sin (\theta_B)} =\frac{AB}{sin (\theta_C)}$

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So

$\frac{BC}{sin (40.2)} = \frac{467.3 }{sin (107.8)}$

=> $BC = 316.8 \ miles$

Also

$\frac{CA}{sin (32)} = \frac{467.3 }{sin (107.8)}$

$CA = 260 .1 \ miles$

Generally the additional flyer miles that Adam will receive if he takes the connecting flight rather than the direct flight is mathematically represented as

$k = [CA +BC] - AB$

=> $k = [260 .1 +316.8]- 467.3$

=> $k =109.6 \ miles$

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