Grayson is 1.25 meters tall. At 12 noon, he measures the length of a tree's shadow to be 22.75 meters. He stands 17.6 meters away from the tree, so that the tip of his shadow meets the tip of the tree's shadow . Find the height of the tree to the nearest hundredth of a meter.

5.52m

Step-by-step explanation:

First, we can see that two similar triangles are formed. One similar triangle is the big triangle, with the base being 22.75m and the height being the tree's height. Another triangle is formed with Grayson's height as the height and the and the difference between where he is standing and the edge of the tree's shadow (the red outline in the picture) .  We know that these are similar triangles because both the bases are parallel, as well as the heights and hypotenuses.

The base length for the triangle with Grayson as the height is 22.75-17.6 = 5.15 .

For similar triangles, we know that the ratios between sides are the same. For example, base₁/base₂ = height₁/height₂ . We can apply this here, making 5.15 base₁ and 1.25 height₁ (note that the same triangle's sides should be on top)

If we make the tree's height t, we thus have

5.15/22.75 = 1.25/t (height 2 is the tree's height

multiply both sides by t to remove one denominator

5.15 * t / 22.75 = 1.25

multiply both sides by 22.75 to remove the other denominator

5.15 * t = 1.25 * 22.75

          = 28.4375

divide both sides by 5.15 to isolate the t

t = 5.52m

y = - 6

step-by-step explanation:

given the point (- 2, y) lies on the graph then it is a solution of it's equation.

substitute x = - 2 into the equation for value of y

y = 3(- 2) = 3 × - 2 = - 6

the point on the graph is (- 2, 6)

22, 28, 34, 40, 46

b)   f(n) = 6n + 16

n = what is the number of the row ( 22 would be row 1) so if we replaced n with 1, we would get 22

c)   the 14th row will have 100 seats

(1)   (2)   (3)   (4)   (5)   (6)   (7)   (8)   (9)   (10) (11) (12) (13) (14)

22 | 28 | 34 | 40 | 46 | 52 | 58 | 64 | 70 | 76 | 82 | 88 | 94 | 100