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pickupchik [31]
2 years ago
6

Which quadratic equation is equivalent itto * (x - 4) ^ 2 - (x - 4) - 6 = 0 ?

Mathematics
1 answer:
kifflom [539]2 years ago
3 0

Answer:

x2 - 9x + 14 = 0

Step-by-step explanation:

x2 - 9x + 14 = 0

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Answer:9000

Step-by-step explanation:

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Help me please I need this done by tomorrow!!!!
Reptile [31]
12) 12, 8

12 - 8 = 4

12 + 8 = 20

Answer: 4 < x <span>< 20

13) 11, 3

11 - 3 = 8

11 + 3 = 14

Answer: 8 </span>< x <span>< 14

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3 years ago
Help me plzzzzzzzzzzzzz​
andreyandreev [35.5K]

Answer:

Q1.

a. three hundred sixty-four thousand eight hundred twenty-five

b. one hundred forty-three

c. nine hundred four thousand three hundred six

d. seven million twenty

e. fifty-six million thirty thousand one hundred twenty-three

f. six thousand twenty-four

g. two hundred four

h. six hundred seventy-four thousand one

i. nine hundred sixty-two thousand three hundred forty-two

Q2.

a. 5000000

b. 1000004

c. 60000203

d. 1311

e. 286

f. 34503271

g. 844536

h. 3748

i. 125064

j. 4605231

5 0
3 years ago
a line segment has one endpoint at the origin (0,0) and one endpoint at (0,6). how long is this segment?
tiny-mole [99]
The segment is 6 units long
(6-0=6)
8 0
3 years ago
How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%
blondinia [14]

The trick is to exploit the difference of squares formula,

a^2-b^2=(a-b)(a+b)

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2

Whatever you do to the denominator, you have to do to the numerator too. So

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2

Expand the numerator:

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}

So we have

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2

But √12 = √(3•4) = 2√3, so

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}

7 0
3 years ago
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