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Q: If A (-2 -4) and B (-8 4) what is the length of?

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Endpoints: A (-2, -4) and B (-8, 4) Length of AB: 10 units

End points: (-2, -4) and (-8, 4) Length of line AB: 10

Using Pythagoras Length AB = √((-8 - 2)² + (4 - -4)²) = √(6² + 8²) = √100 = 10 units.

Using the distance formula length of AB is 6

Using the distance formula the length of ab is 5 units

If a (0 0) and b (8 2) what is the length of ab?

(0,8)2 + (0,2)2 8(2=69 2(2=4 69+4=74√

Use the Pythagorean theorem. a^2 + b^2 = c^2 a = base b = length c = hypotenuse 8^2 + b^2 = 10^2 b^2 = 10^2 - 8^2 b = sqrt(10^2 - 8^2) b = sqrt(100 - 64) b = sqrt(36 b = 6 ------------------the length

a=8 b=6 c=10 answer is 10

8/12 = 4/3b8/12 = 4 x b/33x8/12 = 4xb24/12 = 4b2 = 4b2/4 = 4b/4b = 2/4b = 1/2

4

Length AB is 17 units

12 = -4/2 * -2 + b 12 = -2 * -2 + b 12 = 4 + b 8 = b

first you add 2 to the 8. the 2 crosses out itself so now you have b over 4 equals 10. then you multiply the 4 to the ten and the 4 the 4 crosses out itself and you end up with b= 40

(2,4) and (-4,8) y=mx+b 4=m(2)+b 4=2m+b b=y2-y1 /x2-x1 b = (8-4)/(-4-2) b=4/-2 b=-2 4=2m+-2 6=2m 3=m So your y=mx+b equation would be: y=3x+-2

a + b = 8, a x b = 8 a = 8 - b so 8b - b^2 = 8 ie b^2 - 8b + 8 = 0 Using the quadratic formula gives (8 +/-sqrt(64 - 32)/2 ie (8 +/- 5.66)/2 = 4 +/- 2.83 so the dimensions are 1.17 and 6.83, which sum to 8 and multiply to 7.9911. Near enough?

Let 'a' and 'b' be the length of one side and diagonal of a square. Pythagorus's theorem as applied to a square: a^2 + a^2 = b^2. Substituting a = 2 into the equation, we have b^2 = 2^2 + 2^2 = 8. b = sqrt(8) = 2 * sqrt(2). Q.E.D. ===========================

If you mean: 8/4 then it is 2

Since the area is (length of base)*(height)/2 {call these dimensions B & H) A1 = B*H/2 With dimensions doubled, A2 = (2*B)*(2*H)/2 = 4*B*H/2 = 4*A1. By not simplifying to 2*B*H, it's easier to see that it is four times the original area. It is 4 times because the two length dimensions are multiplied, and 2 * 2 = 4.

-8 + b = 4 add 8 to both sides: b = 12

8/9

The answer is 40. Area of triangle = ba/2 where b = length of the base, a = length of corresponding height. Therefore, 10 X 8/2 = 80/2 = 40.

8=6+b2b=4

Nope. The equation for finding the length of a square's (or right triangle's) diagonal is a^2 + b^2 = c^2. For instance, if a right triangle had one leg (a) with the length of 3, and another leg (b) with the length of 4, the diagonal would have a length of 5. a^2 + b^2 = c^2 3^2 + 4^2 = c^2 9 + 16 = c^2 25 = c^2 5 = c

if the triangle is isosceles right, then the sides' length are : a , a and b. and you have the Pythagorean theorem : a*a +a*a =b*b so if you have the length of the longest side (b) then the two other sides' length are a=b/sqrt(2) = 0,707*b (approx) if you have the length of the 2 same sides (a) then the longest side's length is b=a*sqrt(2) =1,414*a (approx)