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grin007 [14]
3 years ago
6

You are flying your dragon kite. It's connected to 34yd of string. The kite is directly above the edge of the pond. The edge of

the pond is 29yd from where the kite is tied to the ground. How high is the kite above the edge of pond? ( round answer to nearest tenth)
Mathematics
1 answer:
Naily [24]3 years ago
5 0

Answer:

63

Step-by-step explanation:

You might be interested in
Help me pls. I need help with the question
likoan [24]

Answer:

$173.90

Step-by-step explanation:

The most obvious way to answer this question is to multiply $9.40 by 18.5

This would result in 173.90.

But I realize nobody like to deal with decimals in a multiplication problem, so here's the way I would do it (<em>especially if you don't have a calculator!</em>)

The strategy is to break it down!

Multiply $9.40 x 18 hours to start, since this is easier.

9.4 x 18 = 169.2

So, that's $169.20 for 18 hours of work.

We aren't finished, because Abdel worked for 18.5 hours.

So, what half of $9.40?

9.4 / 2

= 4.7

-or-

$4.70

So, knowing that Abdel makes $4.70 in half an hour, lets just add that onto the $169.20 we got earlier.

169.2 + 4.7 = 173.9

-or-

$173.90 for 18.5 hours of work!

8 0
3 years ago
Find dy/dx for y= x^3 ln (cot x)
ICE Princess25 [194]
<h3>Answer</h3>

  \dfrac{dy}{dx} = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)

<h3>Explanation</h3>

By the product rule (d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x), we have

  \begin{aligned}\frac{dy}{dx} &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}

By the chain rule:

  \begin{aligned}\big(\ln (\cot x)\big)' &= \dfrac{1}{\cot x} \cdot (\cot x)' \\ &= \dfrac{1}{\cot x} \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = - \frac{1}{\cos x} \cdot \frac{1}{\sin x} \\&= -\csc(x)\sec(x)\end{aligned}

By the power rule:

  (x^3)' = 3x^2

thus

  \begin{aligned}\frac{dy}{dx} &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}

Nothing to do to simplify any further, other than factoring out x^2.

4 0
4 years ago
If we put money in an investment that increases 14% each year . what would be the multiplier?​
matrenka [14]

Answer:

Step-by-step explanation:

1.14, just like 114%

4 0
3 years ago
Read 2 more answers
WILL GIVE BRAINLIEST
Alex

Answer:

c

Step-by-step explanation:

(3.14)(15*15)(315)=222548

5 0
3 years ago
Plz help me and plz show work 1.
gulaghasi [49]

Answer:

400 800 1200 1600 2000 is anwser

3 0
3 years ago
Read 2 more answers
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