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2 years ago

What is the product (3 x 10') * (2 x 10") in scientific

Mathematics

1 answer:

Mariana [72]2 years ago

5 0

Answer:

600

Step-by-step explanation:

3 x 10 = 30

2 x 10 = 20

30 x 20 = 600

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In the right triangle shown, m∠K =60, and KL=2<br><br> How long is JL?

bagirrra123 [75]

The answer; 90-2 = 88

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3 years ago

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The distance traveled, in feet, of a ball dropped from a tall building is modeled by the equation d(t) = 16t2 where d equals the

nignag [31]

Assuming that by "16t2" you mean $16t^2$ :

First of all, let's compute the position at the two required times:

$d(2)=16\cdot 2^2 = 16*4 = 64$

$d(5)=16\cdot 5^2 = 16*25 = 400$

So, the ball traveled a total distance of $\Delta d = 400-64 = 336$ , during the time $\Delta t = 5-2 = 3$ .

The average rate of change is the ratio between these two:

$\frac{\Delta d}{\Delta t} = \frac{336}{3} = 112$

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3 years ago

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What is the average rate of change of the table below using the interval [‒2, 0]?

Scilla [17]

Answer:

-2x

Step-by-step explanation:

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2 years ago

The length of a rectangle is 1 ft less than double the width, and the area of the rectangle is 21 ft^2. Find the dimensions of t

ICE Princess25 [194]

Let the width be w, then length = 2w - 1

Area = length times width = w(2w - 1) = 21.

Thus, 2w^2 - w = 21 or 2w^2 - w - 21 = 0

Solving the quadratic equation gives that w = 3.5.

Therefore, width is 3.5 ft and length is 2(3.5) - 1 = 6 ft.

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3 years ago

Use chain rule to find dw/dt w=ln(x^2 + y^2 + z^2)^(1/2) , x=sint, y=cost, z=tant

son4ous [18]

The chain rule states that

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac{\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dt}$

Since $\ln(x^2+y^2+z^2)^{1/2}=\dfrac12\ln(x^2+y^2+z^2)$ , you have

$\dfrac{\partial w}{\partial x}=\dfrac x{x^2+y^2+z^2}$
$\dfrac{\partial w}{\partial x}=\dfrac y{x^2+y^2+z^2}$
$\dfrac{\partial w}{\partial z}=\dfrac z{x^2+y^2+z^2}$

and

$\dfrac{\mathrm dx}{\mathrm dt}=\cos t$
$\dfrac{\mathrm dy}{\mathrm dt}=-\sin t$
$\dfrac{\mathrm dz}{\mathrm dt}=\sec^2t$

Also, since $x^2+y^2+z^2=\sin^2t+\cos^2t+\tan^2t=1+\tan^2t=\sec^2t$ , the derivative is

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\sin t\cos t}{\sec^2t}-\dfrac{\sin t\cos t}{\sec^2t}+\dfrac{\tan t\sec^2t}{\sec^2t}$
$\dfrac{\mathrm dw}{\mathrm dt}=\tan t$

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3 years ago