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rjkz [21]
4 years ago
7

Solve each quadratic equation. Show your work.

Mathematics
1 answer:
umka2103 [35]4 years ago
5 0
14. 2x-1 = 0, x+7=0
x = 1/2, x = -7

15. x^2 + 3x - 10 = 0
(x + 5)(x - 2) = 0 
x = -5, x = 2

16. x^2 - 25 = 0
(x-5)(x+5)
x = 5, x  = -5
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Expand 8ps(3qs-2pq+4qs)
shutvik [7]
The answer is 56pqs^2 - 16p^2 sq
5 0
4 years ago
Find the positive solution and please explain how ;4;
valentina_108 [34]
(x + 2)² - 10 = 0 (assuming it's equal to zero)

Expand the binomial: (x + y)² = x² + 2xy + y²
Thus, x² + 4x + 4 - 10 = 0
x² + 4x - 6 = 0

x = \frac{-4 +_- \sqrt{16 + 24}}{2} (by quadratic formula)
= \frac{-4 +_- \sqrt{40}}{2}
= \frac{-4 +_- 2\sqrt{10}}{2}
= -2 + \sqrt{10}, since we're taking the positive solution.
3 0
3 years ago
Find an equation of variation in which y varies inversely as x and y=5 and x=21. Then find the value of y when x=10.
finlep [7]

Answer:

see explanation

Step-by-step explanation:

Given that y varies inversely as x then the equation relating them is

y = \frac{k}{x} ← k is the constant of variation

To find k use the condition y = 5 , x = 21

k = yx = 5 × 21 = 105

y = \frac{105}{x} ← equation of variation

When x = 10, then

y = \frac{105}{10} = 10.5

7 0
3 years ago
An urn contains 13 red balls and 7 blue balls. Suppose that three balls are taken from the urn, one at a time and without replac
d1i1m1o1n [39]

Answer:

0.749

Step-by-step explanation:

The probability that at least one of the three taken balls is blue is the inverse of the probability that none of the three taken balls is blue, aka all 3 of the taken balls are red. The probability of this to happen is

In the first pick: 13/20 chance of this happens

In the 2nd pick: 12/19 chance of this happens

In the 3rd pick: 11/18 chance of this happens

So the probability of picking up all 3 red balls is

\frac{13*12*11}{20*19*18} = \frac{1716}{6840} = 0.251

So the probability of picking up at least 1 blue ball is

1 - 0.251 = 0.749

3 0
3 years ago
Yolanda owns 4 rabbits. She expects the number of rabbits to double every year.
Agata [3.3K]
Double every year
year one
4 times 2
year 2
4 times 2 times 2 aka 4 times 2^2
year 3
4 times 2 times 2 times 2 aka 4 times 2^3
obviously


at year 'n' the number of rabbits will be
4 times 2^n
6 0
4 years ago
Read 2 more answers
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