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Bond [772]
2 years ago
15

PQ has endpoints P(6,-3) and Q(-2,-7). Find the coordinates of the midpoint of PQ.

Mathematics
2 answers:
tester [92]2 years ago
4 0

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Temka [501]2 years ago
3 0

Answer:

Midpoint (2 , -5)

Step-by-step explanation:

RULE: (midpoint)

Consider P(x1,y1)and Q(x2,y2) two points:

the midpoint M(x,y) of the segment PQ is defined like this:

x = \frac{x1+x2}{2} \\\\y=\frac{y1+y2}{2}

In our case

x1 = 6 ; y1 = -3

x2 = -2 ; y2 = -7

then

x = \frac{6+(-2)}{2} =2\\\\y = \frac{(-3)+(-7)}{2} =-5

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3 years ago
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Solve 5x = 2 writing your answer as a fraction in its simplest form
sveticcg [70]

Answer:

x = 2/5

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtract Property of Equality

Step-by-step explanation:

<u>Step 1: Define</u>

5x = 2

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Divide 5 on both sides:                    x = 2/5

<u>Step 3: Check</u>

<em>Plug in x into the original equation to verify it's a solution.</em>

  1. Substitute in <em>x</em>:                     5(2/5) = 2
  2. Multiply:                                2 = 2

Here we see that 2 does indeed equal 2.

∴ x = 2/5 is the solution to the equation.

5 0
3 years ago
A safety officer wants to prove that μ = the average speed of cars driven by a school is less than 25 mph. Suppose that a random
Akimi4 [234]

Answer:

t=\frac{24-25}{\frac{2.2}{\sqrt{14}}}=-1.70    

The degrees of freedom are given by:

df=n-1=14-1=13  

The p value for this case would be given by:

p_v =P(t_{(13)}  

Step-by-step explanation:

Information given

\bar X=24 represent the mean height for the sample  

s=2.2 represent the sample standard deviation

n=14 sample size  

\mu_o =25 represent the value that we want to test

t would represent the statistic

p_v represent the p value for the test

Hypothesis to verify

We want to cehck if the true mean is lees than 25 mph, the system of hypothesis would be:  

Null hypothesis:\mu \geq 25  

Alternative hypothesis:\mu < 25  

The statistic would be given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{24-25}{\frac{2.2}{\sqrt{14}}}=-1.70    

The degrees of freedom are given by:

df=n-1=14-1=13  

The p value for this case would be given by:

p_v =P(t_{(13)}  

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The polar curve $r = 1 + \cos \theta$ is rotated once around the point with polar coordinates $(2,0).$ What is the area of the r
mash [69]

Answer:

Area = -2.3147

Step-by-step explanation:

Given

$r = 1 + \cos \theta$

Required

Determine the area with coordinates (2,0)

The area is represented as:

Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta

Where

$r = 1 + \cos \theta$

and

(a,b) = (2,0)

Substitute values for r, a and b in

Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta

Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)^2} \, d\theta

Expand

Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)(1 + cos\theta)} \, d\theta

Area = \frac{1}{2}\int\limits^0_2 {(1 + 2cos\theta+cos^2\theta} )\, d\theta

By integratin the above, we get:

Area = \frac{1}{2}*\frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{2}[0,2]

Area = \frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{4}[0,2]

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Area = \frac{(cos(0) + 4)sin(0)}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}

Area = \frac{(1 + 4)*0}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}

Area =  - \frac{(cos(2) + 4)sin(2) + 6}{4}

Area =  \frac{-sin(2)(cos(2) + 4) - 6}{4}

Get sin(2) and cos(2) in radians

Area = \frac{-0.9093 * (-0.4161 + 4) - 6}{4}

Area = \frac{-9.2588}{4}

Area = -2.3147

3 0
2 years ago
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