Subtract AB (x+6) from AC (5x-6) to obtain BC
BC should be equal to 4x-12
-3+3r = 12
3r = 12+3
3r = 15
r = 5
Answer:
-23
Step-by-step explanation:
I just did 8+15=23 but made it negative. so you can do
-23+8= -15
Step-by-step explanation:
From Pythagorean theorem, one of the sides can be determined as x^2 + y^2 =8^2
or y = (8^2 - x^2)^(1/2)
we can write the perimeter P as
P = 2x + 2y ---> 20 = 2x + 2(8^2 - x^2)^(1/2)
Dividing by 2, we get
10 = x + (8^2 - x^2)^(1/2)
Move the x to the other side,
10 - x = (8^2 - x^2)^(1/2)
Take the square of both sides to get rid of the radical sign:
(10 - x)^2 = 8^2 - x^2
Move everything to the left and expand the quantity inside the parenthesis:
x^2 + (100 - 20x + x^2) - 64 = 0
2x^2 - 20x + 64 = 0
or
x^2 - 10x + 32 = 0
Now we can see that a = -10 and b = 32
Yes. This equation given:
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" y = (½)x + 4 " ; in point-slope form; also known as: "slope-intercept form" ; is:
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" y = (½)x + 4 " .
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In other words, the equation given is ALREADY written in "point-slope form" ; or, "slope-intercept form".
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Note: An equation that is written in "point-slope form"
(or, "slope-intercept form"), is written in the format of:
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" y = mx + b " ;_________________
in which:_________________
"y" is a single, "stand-alone" variable on the "left-hand side of the equation"; "m" is the coefficient of "x"; also:
"m" is the slope of the line; which is what we want to solve for;
"b" is the "y-intercept"; or more precisely, the value of "x"
(that is; the "x-coordinate") of the point at which "y = 0";
that is, the value of "x" ; or the "x-coordinate" of the point at which
the graph of the equation crosses the "x-axis".
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Note that in our given equation, which is written in "point-slope form" (or, "slope-intercept form" — that is: " y = mx + b " ;
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which is: " y = (½)x + 4 " ;
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we have:
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"y" isolated as "stand-alone" variable on the "left-hand side" of the equation;
m = ½ ;
b = 4 .
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