12. A cyclist travels 40 km at a speed of x km/h. Find the time

Mathematics

2 answers:

GREYUIT [131]1 year ago

7 0

Answer:

Step-by-step explanation:

distance = rate times time, so

distance

time = --------------

rate

Here the distance is 40 km and the rate is x. Then

40 km

time = -------------- = (40/x) hr

x km/hr

If the speed is reduced by 2 km/hr, we get:

40 km 40 km 40

time = ------------------------------- or time = ----------------------- = --------- hr

(x km/hr - 2 km/hr) (x - 2) (km/hr) x - 2

The difference between the times is:

40 40

-------- - --------- = 1 hr Solve this for the original speed, x

x - 2 x

40x - 40x + 80 80

----------------------- = 1 hr or ------------ = 1 or 80 = x^2 - 2x

x(x - 2) x(x - 2)

Rewriting this in standard quadratic form: x^2 - 2x - 80 = 0

Here a = 1, b = -2 and c = -80, and so the discriminant b^2 - 4ac is:

(-2)^2 - 4(1)(-80) = 324

The solutions are:

-(-2) ± √324 2 ± 18

x = -------------------- = ------------ = 1 ± 9, or 10 (discard the other root)

2 2

The original speed was 10 km/hr

Ket [755]1 year ago

4 0

the answer is 10 km/hr.

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part 3. Find the value of the trig function indicated, use Pythagorean theorem to find the third side if you need it.

Nat2105 [25]

Answer: $\bold{9)\ \sin \theta=\dfrac{1}{3}\qquad 10)\ \sin \theta = \dfrac{4}{5}\qquad 11)\ \cos \theta = \dfrac{\sqrt{11}}{6}\qquad 12)\ \tan \theta = \dfrac{17\sqrt2}{26}}$

Step-by-step explanation:

Pythagorean Theorem is: a² + b² = c² , where "c" is the hypotenuse

$9)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{4}{12}\quad \rightarrow \large\boxed{\dfrac{1}{3}}$

Note: 4² + (8√2)² = hypotenuse² → hypotenuse = 12

$10)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{16}{20}\quad \rightarrow \large\boxed{\dfrac{4}{5}}$

Note: 12² + opposite² = 20² → opposite = 16

$11)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{\sqrt{11}}{6}\quad =\large\boxed{\dfrac{\sqrt{11}}{6}}$

Note: adjacent² + 5² = 6² → adjacent = √11

$12)\ \tan \theta=\dfrac{\text{side opposite of}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{17}{13\sqrt2}\quad =\large\boxed{\dfrac{17\sqrt2}{26}}$