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3 years ago

When x pounds of force is applied to one end of a lever that is L feet long, the resulting force y on the other end is

1 answer:

3 0

The question is worded poorly, but it looks like you have a lever in equilibrium, with a force x at a distance d from the fulcrum, and a force y at a distance L - d from the fulcrum. You already have the equilibrium formula for this situation:

xd = y(L - d)

If you know x, y, and d, you can solve for the length L.

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Write the direct variation function given that y varies directly with x, and y = 16 when x = 4.

Otrada [13]

Answer:

Very little information is given, leaving room to multiple possible functions, however i would expect it to be y = 4x

Step-by-step explanation:

x * z = y

4 * z = 16

z = 16/4

z = 4

5 0

3 years ago

The water level in a lake was monitored and was noted to have changed −1 1/5 inches in one year. The next year it was noted to h

pychu [463]

Answer:

the answer is 5 1/9 inches per month (answer)

Step-by-step explanation:

4 0

3 years ago

F(x)=4^-x after a translation 6 units up

Olin [163]

Answer:

Step-by-step explanation:

6 0

2 years ago

A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters

In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

$\Rightarrow y=\sqrt{x^2+100}$

Differentiating with respect to t

$\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}$

$\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}$

Since the car driving towards the intersection at 13 m/s.

so, $\frac{dx}{dt}=-13$

$\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)$

Now

$\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)$

$=\frac{24\times (-13)}{\sqrt{676}}$

$=\frac{24\times (-13)}{26}$

= -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

8 0

3 years ago

What are the zeros of the polynomial equation f(x)=x^2-12x+20?

bagirrra123 [75]

$f(x)=x^2-12x+20\\\\\text{The zeros:}\\\\f(x)=0\to x^2-12x+20=0\\\\x^2-10x-2x+20=0\\\\x(x-10)-2(x-10)=0\\\\(x-10)(x-2)=0\iff x-10=0\ \vee\ x-2=0\\\\\boxed{x=10\ \vee\ x=2}$

8 0

3 years ago