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ad-work [718]
2 years ago
9

Find the equation of the line through the point (-2,-2) that is perpendicular to the line y=2x-1

Mathematics
1 answer:
Lerok [7]2 years ago
5 0

A line perpendicular to y = 2x - 1 must have the slope -1/2 (negative reciprocal of 2).


Thus, the eq'n of the new line is y+2 = (-1/2)(x+2).

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Type the integer that makes the following subtraction sentence true:<br><br> – 7 = -3
Shkiper50 [21]
4 because -7 plus 4 = 3

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What is the least multiple of 6,12 and 15
Jobisdone [24]
The answer would be three 
5 0
3 years ago
Read 2 more answers
Question 1 (Essay Worth 10 points) (01.02 MC) Part A: If (26)x = 1, what is the value of x? Explain your answer. (5 points) Part
katen-ka-za [31]

Answer:

See Explanation

Step-by-step explanation:

The question is not clear. However, I will treat the question as:

(26)x = 1

(50)x = 1

and:

(2^6)^x = 1

(5^0)^x = 1

Solving: (26)x = 1  and  (50)x = 1

(26)x = 1

Divide both sides by 26

x = \frac{1}{26}

(50)x = 1

Divide both sides by 50

x = \frac{1}{50}

Solving (2^6)^x = 1 and (5^0)^x = 1

(2^6)^x = 1

Express 1 as 2^0

(2^6)^x = 2^0

Remove bracket

2^{6x} = 2^0

Cancel out 2

6x = 0

Divide both sides by 6

x = \frac{0}{6}

x = 0

(5^0)^x = 1

Express 1 as 5^0

(5^0)^x = 5^0

Cancel out 5^0

x = 1

8 0
2 years ago
which of the following choices describes the bases of a cylinder a disks be congruent C similar D parallel​
dem82 [27]

Answer: Disc, Congruent, parallel

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Graph the line given by 5x+6y=12 and the circle given by x²+y²=1.Find all solutions to the system of equations.
kenny6666 [7]

Answer:

The system has no real solution.

Step-by-step explanation:

Hi there!

We have the following system of equations:

x² + y² = 1

5x + 6y = 12

The solutions of the system are those (x,y) pairs that satisfy both equations.

Let´s take the second equation and solve it for y:

5x + 6y =12

subtract 5x to both sides of the equation

6y = 12 - 5x

divide by 6 both sides

y = (12- 5x)/6

Now, let´s replace the y in the first equation:

x² + y² = 1

x² + [(12- 5x)/6]² = 1

Apply distributive property

x² + (12 - 5x)²/36 = 1

x² + (12 - 5x)(12 - 5x)/36 = 1

Apply distributive property

x² + (144 - 120x + 25x²)/36 = 1

Apply distributive property

x² + 4 - 10/3 · x + 25/36 · x² = 1

subtract 1 to both sides

61/36 · x² - 10/3 · x + 3 = 0

Using the quadratic formula, let´s find the solutions to this quadratic equation:

a = 61/36

b = -10/3

c = 3

[-b ± √(b² - 4ac)]/2a

[10/3 ± √(100/9 - 4(61/36)(3))]/2(61/36)

(10/3 ± √-83/9)/61/18

This expression has no real solution because √-83/9 is not defined in the set of real numbers. This means that both curves don´t intersect (see attached figure).

Have a nice day!

3 0
3 years ago
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