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Sonbull [250]
3 years ago
15

holly had $5000 in her bank account she withdrew $800 to buy a new bike. what is the percent decreace in the balance of her acco

unt.
Mathematics
1 answer:
Oksana_A [137]3 years ago
6 0
800 was decreased from 5000.
therefore the percentage reduction will be 
(800/5000) * 100
800/50
16%
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To answer this question, you will reflect the original shape to make a new trapezoid at K(2, -1), G (2, -4), H(6, -4), and J (6, -2).  You will keep the y value in each ordered pair and write the opposite x value to show the reflection.

These are all then translated 3 units left to create the new ordered pairs:

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I am on my trigonometry unit in math and don’t know what to do for this. I also really need work shown!
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Answer:

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(make slight changes if directions were given on how to round. This is to the nearest hundredth)

Step-by-step explanation:

Use the Law of Cosines.

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You know two sides and the angle in between. The Law of Cosines will give us the third side.

The third side is the direct path of the airplane. We'll add the two other given sides to find the detour route that they took instead.

Then subtract the smaller number (direct route) from the actual route. See image.

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Factor 12y^2 + 5y - 2 completely.
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dalvyx [7]

Answer:

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Step-by-step explanation:

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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:

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We will let:

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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:

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Answer:

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