All categories

2 years ago

In AGHI, the measure of ZI=90', the measure of ZG-20', and IG = 98 feet. Find

Mathematics

2 answers:

n200080 [17]2 years ago

6 0

Answer:

GH ≈ 104.3 ft

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos20° = $\frac{adjacent}{hypotenuse}$ = $\frac{IG}{GH}$ = $\frac{98}{GH}$

Multiply both sides by GH

GH × cos20° = 98 ( divide both sides by cos20° )

GH = $\frac{98}{cos20}$ ≈ 104.3 ft ( to the nearest tenth )

ohaa [14]2 years ago

6 0

Answer:

GH = 104.3 ft

Step-by-step explanation:

You have a right triangle, so you can use a trig ratio: sine, cosine, or tangent.

You are given the measure of angle G and the length of leg IG. Side GH is the hypotenuse. For angle G, side IG is the adjacent leg. The trig ratio that relates the adjacent leg and the hypotenuse is the cosine.

$cos A = \dfrac{adj}{hyp}$

$cos G = \dfrac{IG}{GH}$

$cos 20^\circ = \dfrac{98}{x}$

$xcos 20^\circ = 98$

$x = \dfrac{98}{cos 20^\circ}$

$x = \dfrac{98}{0.9397}$

$x = 104.3$

Answer: GH = 104.3 ft

You might be interested in

Will a 9m long planks fit into a square room of side 7m? please I really needed help with this one

Darya [45]

No because if it is a square all the sides are 7m long so a 9m plank will not fit. Hope this helps.

5 0

2 years ago

When a national sample of heights of kindergarten children was? taken, a student was told that he was in the 20th percentile.

Nutka1998 [239]

Answer:

this question doesnt make sense as its like half of it

6 0

2 years ago

1. Given a regular octagon, find the measures of the angles formed by (a) two consecutive radii and (b) a

Fed [463]

A ) The angle formed by 2 consecutive radii:
Ф= 360° : 8 = 45°
b ) The angle formed by a radius and a side of a polygon:
α = (180° - 45°): 2 = 135° : 2 = 67.5°
Answer:
B ) 45°, 67.5°

4 0

2 years ago

Use the given transformation to evaluate the given integral, where r is the triangular region with vertices (0, 0), (8, 1), and

Jlenok [28]

We first obtain the equation of the lines bounding R.

For the line with points (0, 0) and (8, 1), the equation is given by:

$\frac{y}{x} = \frac{1}{8} \\ \\ \Rightarrow x=8y \\ \\ \Rightarrow8u+v=8(u+8v)=8u+64v \\ \\ \Rightarrow v=0$

For the line with points (0, 0) and (1, 8), the equation is given by:

$\frac{y}{x} = \frac{8}{1} \\ \\ \Rightarrow y=8x \\ \\ \Rightarrow u+8v=8(8u+v)=64u+8v \\ \\ \Rightarrow u=0$

For the line with points (8, 1) and (1, 8), the equation is given by:

$\frac{y-1}{x-8} = \frac{8-1}{1-8} = \frac{7}{-7} =-1 \\ \\ \Rightarrow y-1=-x+8 \\ \\ \Rightarrow y=-x+9 \\ \\ \Rightarrow u+8v=-8u-v+9 \\ \\ \Rightarrow u=1-v$

The Jacobian determinant is given by

$\left|\begin{array}{cc} \frac{\partial x}{\partial u} &\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{array}\right| = \left|\begin{array}{cc} 8 &1\\1&8\end{array}\right| \\ \\ =64-1=63$

The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v

Therefore, the integration is given by:

$63 \int\limits^1_0 \int\limits^{1}_0 {(5u-23v)} \, dudv =63 \int\limits^1_0\left[\frac{5}{2}u^2-23uv\right]^{1}_0 \\ \\ =63\int\limits^1_0(\frac{5}{2}-23v)dv=63\left[\frac{5}{2}v-\frac{23}{2}v^2\right]^1_0=63\left(\frac{5}{2}-\frac{23}{2}\right) \\ \\ =63(-9)=|-576|=576$

6 0

2 years ago

Store A

olga2289 [7]

Answer:

Step-by-step explanation:

3 0

2 years ago