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3 years ago

8

Simon draws a rectangle with two sides that are 2 inches long and two sides that are 3 inches long.Chaz draws a figure with the

same measurements,but it is not a rectangle.What figure could it be.

1 answer:

7nadin3 [17]3 years ago

3 0

The other shape might be a trapezoid.

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On the map of a town, the library is located at (–8, 8), and the seafood market is located at (–8, 2). Mary walks from the libra

sertanlavr [38]

6 units. have a great day!!!!

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4 years ago

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Owen is signing up for a gym membership with a one-time fee to join and then a

Katena32 [7]

The cost of the monthly fee is 50$.. if the joining fee is 100.. take 200 and subtract that fee since it is included in the total. You’re left with 100. Note, the 200 is calculated for a 2 month period, so split the remaining 100 ,from the total 200, into 2. You’re left with 50 each month.

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3 years ago

Find derivative problem<br> Find B’(6)

dalvyx [7]

Answer:

$B^\prime(6) \approx -28.17$

Step-by-step explanation:

We have:

$\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)$

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

$\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]$

We can move the constant outside:

$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

$(uv)^\prime=u^\prime v+u v^\prime$

We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$

Therefore:

$\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]$

By simplification:

$\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17$

So, the slope of the tangent line to the point (6, B(6)) is -28.17.

5 0

3 years ago

I need help with d 1 to d 3. I don't understand what to do.

erik [133]

What????

You just add 1 +3 =4

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3 years ago

Solve the inequality.<br><br> 2(4x – 3) ≥ –3(3x) + 5x?

slega [8]

2(4x-3) ≥ - 3(3x) + 5x

2* 4x - 2*3 ≥ - 9x + 5x

8 x - 6 ≥ -4x

8 x + 4 x ≥ 6

12 x ≥ 6

x = 6/12 divide 6/6 = 1 and 12 / 6 = 2

x = 1/2

hope this helps!

4 0

3 years ago