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Harrizon [31]
3 years ago
9

If P is the circumcenter △ A B C, and AD =3x - 11, DB =5x - 29, PC =18, find DP.

Mathematics
2 answers:
Lunna [17]3 years ago
6 0

Answer:

8.2

Step-by-step explanation:

1. Since AD is equal to DB, we will need to set 3x-11 equal to 5x-29. When you solve the equation you will get 9 for x.

2. Plug the 9 in for x in DB (you could plug it in for AD instead if you want since DB and AD are equal to each other). You should get 16.

3. Take a look at the smaller triangle DBP which is part of triangle ABC. We will need to use the Pythagorean Theorem. Square the legs of the triangle and set it equal to the hypotenuse squared. One of the legs, DP, is unknown so we will just have it as x. The other leg is DB, which is 16.

* PB is the hypotenuse and has a value of 18. Remember that P is the circumcenter, which means it is equidistant to each vertex and each angle bisector is equal to each other. As a result PC is equal to PB which happens to be 18.

You should have it something like this:

16^2 + x^2 = 18^2

Now do all the squaring and stuff and you should eventually get x^2 = 68

4. Square root x^2 = 68 and you will get a long decimal. You will need to round it to the tenths place.

5. After rounding you will finally get 8.2

Basile [38]3 years ago
5 0

Answer:

15.6

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

8 0
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grin007 [14]

Answer:

a = 58

Step-by-step explanation:

90-32=58

HOPE THIS HELPS :)

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2 years ago
Which of the following tables represents a proportional relationship?
algol13

Answer:

You’re right!

Step-by-step explanation:

It IS 3, heres why:

it follows the same rule each time:

2X=Y

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thats proportionate! :D

i hope this helps!

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