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LenaWriter [7]
3 years ago
9

The first two steps in determining the solution set of the system of equations, y = x2 – 6x + 12 and y = 2x – 4, algebraically a

re shown in the table.Which represents the solution(s) of this system of equations? (4, 4) (–4, –12) (4, 4) and (–4, 12) (–4, 4) and (4, 12)
Mathematics
2 answers:
olchik [2.2K]3 years ago
3 0

Answer:

y = 2x - 4....so sub in 2x - 4 for x in the other equation

y = x^2 - 6x + 12

2x - 4 = x^2 - 6x + 12

x^2 - 6x - 2x + 12 + 4 = 0

x^2 - 8x + 16 = 0

(x - 4)(x - 4) = 0

x - 4 = 0

x = 4

x - 4 = 0

x = 4

solution is (4,4)

(2, −1) and (−4, 17)

Step-by-step explanation:

DedPeter [7]3 years ago
3 0

Answer:

wrong up top

Step-by-step explanation:

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46,100

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Can anyone teach mee..................... ​
andrew11 [14]

Answer:

(i) The name of the part of the circle, OQ is a radius

(ii) The radius of the sector QOR is 21 cm

Step-by-step explanation:

The given figure is a sector of the circle O

∵ Any sector of a circle formed from 2 radii and an arc

∴ OQ is a radius

(i) The name of the part of the circle, OQ is a radius

The rule of the length of an arc of a circle is L = \frac{\alpha }{360} × 2 π r, where

  • α is the angle of the sector
  • r is the radius of the circle

∵ The length of the arc QR is 22 cm

∴ L = 22

∵ The measure of the angle of the arc is 60°

∴ α = 60°

∵ π = \frac{22}{7}

→ Substitute them in the rule above

∵ 22 = \frac{60}{360} × 2 × \frac{22}{7} × r

∴ 22 = \frac{22}{21} r

→ Divide both sides by  \frac{22}{21}

∴ 21 = r

(ii) The radius of the sector QOR is 21 cm

5 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}

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3 years ago
Find the slope of the line that passes through (8, 4) and (2, 9).
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The answers is not -5/6
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