You can take it apart. There are a top and bottom (both the same) right triangle. So you can find the area of that by multiplying 8*6 and divide by two. Then multiply by two because there are 2 triangles.
You are left with three rectangular sides: One 10x10, one 10x6, and one 10x8.
So your whole equation looks like this: A = 2[(8*6)/2]+(10*10)+(10*6)+(10*8)
Let's solve this system of equations through substitution.
We have these two equations.
-7x-2y=14
6x+6y=18
Now let divide the second equation by 6.
6x+6y=18 ----> x+y=3
Next, let us move y to the right side of the equation.
x+y=3 -------> x=3-y (x equals 3-y)
Because we found out that what x is in terms of y, we can input that in for every instance of x in this equation below.
-7x-2y=14 becomes -7(3-y)-2y=14 (Why? Because x equals 3-y!)
We have a one variable equation now and can solve for y.
-7(3-y)-2y=14
-21+7y-2y=14
5y=35
y=7
Plug in 7 for y in any equation to find x.
x+y=3
x+7=3
x=-4
answer: x=-4, y=7
Answer:
Step-by-step explanation:
a. sin^-1(sin theta) = theta
(1/sin)(sin theta) = theta
<em>the sines cross out</em> theta = theta
b. cos(cos^-1x) = x
cos(x/cos) = x
<em>the cosines cross out </em>x = x
Answer:
x^2 + y^2 = 4
Step-by-step explanation:
The center-radius form (formally called the standard form) of a circle is
(x-h)^2+(y-k)^2=r^2 where (h,k) is the center and r is the radius.
So if we replace (h,k) with (0,0) since the center is the origin and r with 2 since the radius is 2 we get:
(x-0)^2+(y-0)^2=2^2
Let's simplify:
x^2 + y^2 = 4