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vredina [299]
3 years ago
13

What is the slope of the line shown below?

Mathematics
1 answer:
strojnjashka [21]3 years ago
6 0

Because we're given two points on the line, we can apply the slope formula to find the answer:

\frac{y.2 - y.1}{x.2 - x.1} = m

-4 - 8 / 2 - (-1)

-12 / 3

-4

<h3>The slope of the line is -4.</h3>
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Fiesta28 [93]
Here is how you find the answer. 
Do remember that the term Bisect means to cut it in half.
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<span>5x = 3x + 10
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</span>Remember bisect? so 25 x 2 so the final answer is 50.
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a circle has a center (3,5) and a diameter AB. The coordinates of A are (-4,6). what are the coordinates of B?
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r= \sqrt{(3-(-4))^2+(5-6)^2}&#10;\\r =  \sqrt{7^2+(-1)^2}&#10;\\r= \sqrt{50} &#10;\\  r=5 \sqrt{2}

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Point B must lie on a line AC, where C is a center of a circle.
Find equation of line AC.
A(–4, 6), C(3, 5)

y-y_1= \frac{y_2-y_1}{x_2-x_1}(x-x_1)&#10;\\y-6= \frac{5-6}{3-(-4)}  (x-(-4))&#10;\\y-6=- \frac{1}{7} (x+4)&#10;\\7y-42=-x-4&#10;\\x+7y-38=0 

The distance from B(x, y) to C(3, 5) is 5√2.
\sqrt{(x-3)^2+(y-5)^2}=5 \sqrt{2} &#10; \\(x-3)^2+(y-5)^2=(5 \sqrt{2})^2 &#10;\\(x-3)^2+(y-5)^2=50

Solve system of equations.
x+7y-38=0&#10;\\(x-3)^2+(y-5)^2=50&#10;\\&#10;\\x=38-7y&#10;\\(38-7y-3)^2+(y-5)^2=50&#10;\\(35-7y)^2+(y-5)^2=50&#10;\\1225-490y+49y^2+y^2-10y+25-50=0&#10;\\50y^2-500y+1200=0&#10;\\y^2-10y+24=0&#10;\\y^2-6y-4y+24=0&#10;\\y(y-6)-4(y-6)=0 \\(y-6)(y-4)=0 \\y_1=6,y_2=4&#10;

x_1=38-7y_1=38-7 \times 6 = 38-42=-4&#10;\\x_2=38-7y_2=38-7 \times 4 = 38-28=10&#10;

Point B could have coordinates
\\&#10;\\(x_1,y_1)=(-4,6),(x_2,y_2)=(10,4)&#10;

But (–4, 6) are the coordinates of point A.
Therefore, point B has coordinates (10,4).
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3 years ago
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