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abruzzese [7]
3 years ago
13

What is the midpoint of the segment whose endpoints are (17,1) and (-9,3)?

Mathematics
2 answers:
bogdanovich [222]3 years ago
5 0
Use the midpoint formula, giving you the point (4,2)
Len [333]3 years ago
5 0

Answer:

(4, 2)

Step-by-step explanation:

To find the midpoints of endpoints, you need to find the average of the numbers.

17+(-9)=8/2=4

1+3=4/2=2

So, the endpoints are (4,2)

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Construct<span> a perpendicular from the </span>incenter<span> to one side of the triangle to locate the exact radius. 3. place compass point at the </span>incenter<span> and measure from the center to the point where the perpendicular crosses the side of the triangle</span>
7 0
3 years ago
Which sentence explains the correct first step in the solution of this equation? <br> 4(3+x)=9
QveST [7]

Answer:

Multiply 4(3+x)

Step-by-step explanation:

First:  4*3=12 so, 12+3x=9

12+3x=9

-12

3x=-3

divide by 3

Solution: x=-1

6 0
4 years ago
I need help with c please.
STALIN [3.7K]

Answer:

||w|| = 6

Step-by-step explanation:

||w|| = modulus or magnitude of vector w

Since formula to get the modulus of any vector = \sqrt{\text{(x-component)}^2+\text{(y-component)}^2}

Vector w = <-6, 0>

x-component of vector w = (-6)

y-component of vector w = 0

Therefore, ||w|| = \sqrt{(-6)^2)+(0)^2}

                         = 6

3 0
3 years ago
Evaluate rg + 2d when r = 3, g = 5, and d = 2
qwelly [4]

Answer:

rg+2d= 12

Step-by-step explanation:

(3+5)+2(2)

8+4=12

3 0
3 years ago
Find all solutions to the equation in the interval [0, 2π).<br><br> cos 4x - cos 2x = 0
Andrew [12]
Solve cos(4x)-cos(2x)=0  &forall; 0<=x<=2pi  ..............(0)

Normal solution:
1. use the double angle formula to decompose, and recall cos^2(x)+sin^2(x)=1
cos(4x)=cos^2(2x)-sin^2(2x)=2cos^2(2x)-1    .................(1)
2. substitute (1) in (0)
2cos^2(2x)-1-cos(2x)=0
3. substitute u=cos(2x)
2u^2-u-1=0
4. Solve for x
factor
(u-1)(u+1/2)=0
=> u=1 or u=-1/2
However, since cos(x) is an even function, so solutions to
{cos(2x)=1, cos(-2x)=1, cos(2x)=-1/2 and cos(-2x)}  ...........(2)
are all solutions.
5. The cosine function is symmetrical about pi, therefore
cos(-2x)=cos(2*pi-2x), 
solution (2) above becomes
{cos(2x)=1, cos(2pi-2x)=1, cos(2x)=-1/2, cos(2pi-2x)=-1/2}
6. Solve each case
cos(2x)=1 => x=0
cos(2pi-2x)=1 => cos(2pi-0)=1 => x=pi
cos(2x)=-1/2 => 2x=2pi/3 or 2x=4pi/3 => x=pi/3 or 2pi/3
cos(2pi-2x)=-1/2 => 2pi-2x=2pi/3 or 2pi-2x=4pi/3 => x=2pi/3 or x=4pi/3
Summing up,
x={0,pi/3, 2pi/3, pi, 4pi/3}
3 0
3 years ago
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