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Yuliya22 [10]
3 years ago
5

Solve for b. Put your answer in simplest radical form. Show Steps plz! A = 4πbc + πb^2

Mathematics
1 answer:
sladkih [1.3K]3 years ago
4 0

Answer:

b = -2c ± [√(4π²c² + πA)]/π

Step-by-step explanation:

A = 4πbc + πb^2

A = 4πbc + πb²

πb² + 4πbc - A = 0

Using the quadratic formula to solve this quadratic equation.

The quadratic formula for the quadratic equation, pb² + qb + r = 0, is given as

b = [-q ± √(q² - 4pr)] ÷ 2p

Comparing

πb² + 4πbc - A = 0 with pb² + qb + r = 0,

p = π

q = 4πc

r = -A

b = [-q ± √(q² - 4pr)] ÷ 2p

b = {-4πc ± √[(4πc)² - 4(π)(-A)]} ÷ 2π

b = {-4πc ± √[16π²c² + 4πA]} ÷ 2π

b = (-4πc/2π) ± {√[16π²c² + 4πA] ÷ 2π}

b = -2c ± [√(4π²c² + πA)]/π

Hope this Helps!!!

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Answer:

A. 57.6

Step-by-step explanation:

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What is the solution to the system of equations? Use the substitution method. {y=2x+43x−6y=3
garri49 [273]
We have that
<span>y=2x+4--------> equation 1
3x−6y=3-------> equation 2

step 1
</span>I substitute the value of y in equation 1 for the value of y in equation 2<span>
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step 2
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the answer is 
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Cerrena [4.2K]

Step-by-step explanation:

Since it remains only 1 sweet, we can subtract it from the total and get the amount of sweets distributed (=1024).

As all the sweets are distributed equally, we must divide the number of distributed sweets by all its dividers (excluding 1024 and 1, we'll see later why):

1) 512 => 2 partecipants

2) 256 => 4 partecipants

3) 128 => 8 partecipants

4) 64 => 16 partecipants

5) 32 => 32 partecipants

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10) 2 => 512 partecipants

The number on the left represents the number of sweets given to the partecipants, and on the right we have the number of the partecipants. Note that all the numbers on the left are dividers of 1024.

Why excluding 1 and 1024? Because the problem tells us that there remains 1 sweet. If there was 1 sweet for every partecipant, the number of partecipants would be 1025, but that's not possible as there remains 1 sweet. If it was 1024, it wouldn't work as well because the sweets are 1025 and if 1 is not distributed it goes again against the problem that says all sweets are equally distributed.

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d) There is one positive real root, at x=13. A graphical solution works well.

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