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1 year ago

Find sin θ if θ is in Quadrant III and tan θ = -1/5

Mathematics

1 answer:

Alborosie1 year ago

4 0

Answer:= - sin( square root 8 over 3)sin(1/4)

Step-by-step explanation:

So I did (square root 8/3)(square root 15/4)+sin(1/4)cos(-1/3) divide that by (-1/3)(square root 15 over 4) - sin( square root 8 over 3)sin(1/4)

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Sloan [31]

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3 years ago

A. Write an expression to represent the difference between the two points shown on the number line.

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3 years ago

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Step-by-step explanation:

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2 years ago

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The triangles are similar. Solve for the missing segment.

PIT_PIT [208]

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Since the triangles are similar then the ratios of corresponding sides are equal, that is

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2 years ago

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A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2.

IRISSAK [1]

Answer:

80 feet

Step-by-step explanation:

Given:

Initial speed of the car ( $v_0$ ) = 40 ft/sec

Deceleration of the car ( $\frac{dv}{dt}$ ) = -10 ft/sec²

Final speed of the car ( $v_x$ ) = 0 ft/sec

Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.

Now, deceleration means rate of decrease of velocity.

So, $\frac{dv}{dt}=-10\ ft/sec^2$

Negative sign means the velocity is decreasing with time.

Now, $\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})$ using chain rule of differentiation. Therefore,

$\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx$

Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,

$\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft$

Therefore, the car travels a distance of 80 feet before stopping.

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2 years ago