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Sladkaya [172]
3 years ago
5

Let g(x) = |x + 1|, solve the equation g(x) = 2x + 4

Mathematics
1 answer:
hammer [34]3 years ago
5 0

Answer:

Step-by-step explanation:

g(x)= -x-1 or g(x) = x+1

if g(x) = -x-1

-x-1=2x+4

-x=2x+5

-x/2=x+5

-x/2 -x=5

-x/2-2x/2=5

-x-2x=5

-3x=5

x=5/-3

if g(x) = x+1

x+1=2x+4

x=2x+3

x/2=x+3

x/2-x=3

x/2-2x/2=3

x-2x=3

-x=3

x=3

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Solve by using square roots. ​ 6x^2−27=39
Bingel [31]

Answer:

Simplifying

x = 11

Step-by-step explanation:

Simplifying

6x + -27 = 39

Reorder the terms:

-27 + 6x = 39

Solving

-27 + 6x = 39

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '27' to each side of the equation.

-27 + 27 + 6x = 39 + 27

Combine like terms: -27 + 27 = 0

0 + 6x = 39 + 27

6x = 39 + 27

Combine like terms: 39 + 27 = 66

6x = 66

Divide each side by '6'.

x = 11

Simplifying

x = 11

6 0
3 years ago
What’s the slope of a line that passes through the points (-3,7) and (-3, 4)?
Dmitry_Shevchenko [17]

Answer: The slope should be infinity.

Step-by-step explanation: I hope this helps you!

4 0
4 years ago
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Given that MQL =180° and XQR = 180°, which equation could be used to solve problems involving the relationships
Ganezh [65]

Answer:

Option A is the right answer.

146-4b = 122-1b

Reason: vertical angles

8 0
3 years ago
2. Consider the circle x^2−4x+y^2+10y+13=0. There are two lines tangent to this circle having a slope of 2/3.
likoan [24]

Answer:

a) The coordinates are (0.431, -2.646) and (3.568,-7.354)

b) The tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

Step-by-step explanation:

First, lets complete squares, by taking for each cordinate the square of a linear expression

0= x²−4x+y²+10y+13 = (x-2)²+ 4  + (y+5)²-25+13 = (x-2)²+(y+5)² - 8

Hence (x-2)² + (y+5)² = 8

Lets put y in function of x. We should obtain 2 functions f and g that represent the circle.

(y+5)² = 8 - (x-2)² = -x² + 4x + 4

y = ^+_- \sqrt{-x^2+4x+4} -5

Thus

f(x) = \sqrt{-x^2+4x+4} - 5

g(x) = -\sqrt{-x^2+4x+4} - 5

Lets find the derivate of each function and the points in which they reach the value 2/3. In those points the tangent line will have a slope of 2/3.

We may just find the values of x which the derivate of f is either 2/3 or -2/3.

\frac{-x+2}{\sqrt{-x^2+4x+4}} = ^+_-\frac{2}{3} \Leftrightarrow \frac{x^2-4x+4}{-x^2+4x+4} = \frac{4}{9} \Leftrightarrow 9(x^2-4x+4) = 4(-x^2+4x+4) \\\Leftrightarrow 13x^2-52x+20 = 0

The quadratic has roots

\frac{52 ^+_-\sqrt{1664}}{26}

one root is 3.568, which corresponds with g, and the other root is 0.431, corresponding to f.

Also g(3.568) = - 7.354 and f(0.431) = -2.646

This means that the coordinates are (0.431 , -2.646), (3.568 , -7.354) and the tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

5 0
3 years ago
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mel-nik [20]

the correct answer is a, because a vertical angle is made of 2 intercepting lines.

3 0
3 years ago
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