H = 2sqrt2
D = 6sqrt3
In a 45,45,90 triangle the hypotenuse is (x)sqrt2 while the side lengths are equivalent being a single value x. Therefore, when given the hypotenuse and solving for the leg, divide 4 by sqrt2 to get 4sqrt2/2 which simplifies to 2sqrt2 when the denominator and numerator cancel.
In a 30,60,90 triangle the short leg x is across from the 30 degree angle meaning the angle across from the 60 degree angle is x times the sqrt of 3. Therefore the long leg is 6sqrt3
Do you have to keep going or only just one more number it seems to be going by 2’s 4,6,8,10,12 but I’m a little confused since it says it’s 9th grade work
Answer:
See below.
Step-by-step explanation:
a.
O is the origin, so its abscissa is 0. O: 0
The only point 4 units from O is A, so A: 4
Point C has a positive abscissa like A.
Points B and C have opposite abscissas. The only opposites are -3 and 3, but C is positive, and B is negative.
B: -3
C: 3
D: -4.5
E: -6
M = (4 + (-6))/2 = -1. M: -1
b.
OB = |0 - (-3)| = |3| = 3
DA = |-4.5 - 4| = |-8.5| = 8.5
The answer is
and
. A picture is attached to explain what alternate exterior angles are.
Y - 2 = -5(x-4)
y - 2 = -5x + 20
y = -5x +22
