Yeah.. baby, do the SOH CAH TOA with the hula hoop
![\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad \qquad % cosine cos(\theta)=\cfrac{adjacent}{hypotenuse} \\ \quad \\ % tangent tan(\theta)=\cfrac{opposite}{adjacent} ](https://tex.z-dn.net/?f=%5Cbf%20%0Asin%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Bhypotenuse%7D%0A%5Cqquad%20%5Cqquad%20%0A%25%20cosine%0Acos%28%5Ctheta%29%3D%5Ccfrac%7Badjacent%7D%7Bhypotenuse%7D%0A%0A%5C%5C%20%5Cquad%20%5C%5C%0A%25%20tangent%0Atan%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Badjacent%7D%0A)
so.. hmmm we have in the picture, an angle, and two sides,
actually, is the "adjacent" side, and the "opposite"
so.. which of those ratios give us only
the angle
the adjacent side
the opposite side?
ahhha!, is Ms tangent
so
![\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(42^o)=\cfrac{h}{100}](https://tex.z-dn.net/?f=%5Cbf%20tan%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Badjacent%7D%5Cimplies%20tan%2842%5Eo%29%3D%5Ccfrac%7Bh%7D%7B100%7D)
solve for "h"
It would take her 45 mins to grade 2 stacks or 3/4 of an hour.
The answer is 28sqinches Why The Surface Area of a pyramid is A=L×W×Have since the base is l×with combined it's now SA=BA×H
SA=18×16
SA=288sqinches.
Answer:
First term = -41.
Common difference = -15.
Step-by-step explanation:
nth term: an = a1 + (n - 1)d where a = first term and d = the common difference.
-671 = a1 + (43 - 1)d
-806 = a1 +(52 - 1)d
-671 = a1 + 42d
-806 = a1 + 51d
Subtracting ( to eliminate a1):
-671 - (-806) = 42d - 51d
-9d = 135
d = -15
Substitute for d in the first equation:
-671 = a1 + 42*-15
-671 = a1 - 630
a1 = -671 + 630 = -41.
Answer:
13.4±2.132(1.02/5)
Step-by-step explanation:
Khan Academy