Answer:
49.62
Step-by-step explanation:
You can start by dividing the shape into 2 separate rectangles to make calculating the area easier.
9.3 x 2.4 = 22.32
10.5 x 2.6 = 27.3
Then, add the areas of the two separate rectangles to find the total area of the shape.
22.32 + 27.3 = 49.62. Hope that helped
To solve the question we shall use the formula for the range given by:
Horizontal range, R=[v²sin 2θ]/g
plugging in our values we get:
500=[160²×sin 2θ]/10
5000=160²×sin 2θ
0.1953=sin 2θ
thus:
arcsin 0.1953=2θ
11.263=2θ
hence:
θ=5.6315°~5.63
Radius is half of diameter.
Circumference can be found using the equation C=2(pi)r or C=(pi)d.
r=radius, d=diameter
(pi) is about 3.14 or 22/7
Hello :
<span> formula for the linear function with slope -9 and x-intercept 2 is :
y - 0 = -9(x-2)
y= -9x +18
</span>
Answer:
There are two choices for angle Y: 
for 
, 
for 
.
Step-by-step explanation:
There are mistakes in the statement, correct form is now described:
<em>In triangle XYZ, measure of angle X = 49°, XY = 18 and YZ = 14. Find the measure of angle Y:</em>
The line segment XY is opposite to angle Z and the line segment YZ is opposite to angle X. We can determine the length of the line segment XZ by the Law of Cosine:
(1)
If we know that 
, 
and 
, then we have the following second order polynomial:
(2)
By the Quadratic Formula we have the following result:
There are two possible triangles, we can determine the value of angle Y for each by the Law of Cosine again:
1)
![Y = \cos^{-1}\left[\frac{18^{2}+14^{2}-15.193^{2}}{2\cdot (18)\cdot (14)} \right]](https://tex.z-dn.net/?f=Y%20%3D%20%5Ccos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B18%5E%7B2%7D%2B14%5E%7B2%7D-15.193%5E%7B2%7D%7D%7B2%5Ccdot%20%2818%29%5Ccdot%20%2814%29%7D%20%5Cright%5D)
2)
![Y = \cos^{-1}\left[\frac{18^{2}+14^{2}-8.424^{2}}{2\cdot (18)\cdot (14)} \right]](https://tex.z-dn.net/?f=Y%20%3D%20%5Ccos%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B18%5E%7B2%7D%2B14%5E%7B2%7D-8.424%5E%7B2%7D%7D%7B2%5Ccdot%20%2818%29%5Ccdot%20%2814%29%7D%20%5Cright%5D)
There are two choices for angle Y: for , for .
