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Gennadij [26K]
3 years ago
15

How to find the other endpoint of a line when only given the first endpoint and the midpoint?

Mathematics
1 answer:
Schach [20]3 years ago
6 0
If we have a line, and the endpoints are(x₁,y₁) and (x₂,y₂), the midpoint will be:( (x₁+x₂)/2 , (y₁+y₂)/2 );

Data:
Endpoint₁: (2,-3)
Midpoint: (6,6)

Therefore:

(2+x₂)/2=6
2+x₂=12
x₂=12-2
x₂=10

(-3+y₂)/2=6
-3+y₂=12
y₂=12+3
y₂=15

Answer; the other endpoint of a line will be (10,15)


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A=s^2

A=6.3^2

(A=6.3x6.3)


And your answer would be:


A=39.69cm^2

4 0
3 years ago
3/4 of analisa's friends came to her birthday party. express this fraction as a percant and a decimal.
icang [17]

Answer:

0.75 And 75%

Step-by-step explanation:

If 3/4 came then 25, 50, 75, 100 what is the 3rd number of of those 4

8 0
3 years ago
Hey can someone help find the area of these two pls ty!
Hitman42 [59]

Step-by-step explanation:

I think the answer for the 3) is S=(a×ha)÷2= (7×4.2)÷2= 29,4 ÷2=14.7

And for the 5) one im really not sure.

5 0
2 years ago
The interior angle sum of a convex polygon is 330, how many sides does the polygon have?
MissTica

let the number of sides be n.

Thus,

(n-2)x180=330

n-2=1.83

n=3.83

Rounding off we have, n=4.

Thus the total number of side is 4.

4 0
1 year ago
A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 31 ft/s. (a) A
julsineya [31]

Answer:

a) -13.9 ft/s

b) 13.9 ft/s

Step-by-step explanation:

a) The rate of his distance from the second base when he is halfway to first base can be found by differentiating the following Pythagorean theorem equation respect t:

D^{2} = (90 - x)^{2} + 90^{2}   (1)

\frac{d(D^{2})}{dt} = \frac{d(90 - x)^{2} + 90^{2})}{dt}

2D\frac{d(D)}{dt} = \frac{d((90 - x)^{2})}{dt}  

D\frac{d(D)}{dt} = -(90 - x) \frac{dx}{dt}   (2)

Since:

D = \sqrt{(90 -x)^{2} + 90^{2}}

When x = 45 (the batter is halfway to first base), D is:

D = \sqrt{(90 - 45)^{2} + 90^{2}} = 100. 62

Now, by introducing D = 100.62, x = 45 and dx/dt = 31 into equation (2) we have:

100.62 \frac{d(D)}{dt} = -(90 - 45)*31          

\frac{d(D)}{dt} = -\frac{(90 - 45)*31}{100.62} = -13.9 ft/s

Hence, the rate of his distance from second base decreasing when he is halfway to first base is -13.9 ft/s.

b) The rate of his distance from third base increasing at the same moment is given by differentiating the folowing Pythagorean theorem equation respect t:

D^{2} = 90^{2} + x^{2}  

\frac{d(D^{2})}{dt} = \frac{d(90^{2} + x^{2})}{dt}

D\frac{dD}{dt} = x\frac{dx}{dt}   (3)

We have that D is:

D = \sqrt{x^{2} + 90^{2}} = \sqrt{(45)^{2} + 90^{2}} = 100.63

By entering x = 45, dx/dt = 31 and D = 100.63 into equation (3) we have:

\frac{dD}{dt} = \frac{45*31}{100.63} = 13.9 ft/s

Therefore, the rate of the batter when he is from third base increasing at the same moment is 13.9 ft/s.

I hope it helps you!

4 0
3 years ago
Read 2 more answers
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