Answer:
A = 1100cm^2
Step by step Explanation:
given the dimensions of width and length of the picture are x+20 by 2x-10 and the frame is a constant 5 cm wider from the edge of the picture to the frame, than the area of the frame is defined as (x+30)(2x)-(x+20)(2x-10) = 30x+200.
If the width is equal to the length which I assume is true if the width is constant.
than x+30=2x, which means x = 30.
if this is true than 30(30)+200 = 1100cm^2
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Step-by-step explanation:
Answer:
or ![(-\infty, 47/6]](https://tex.z-dn.net/?f=%28-%5Cinfty%2C%2047%2F6%5D)
Step-by-step explanation:
Note that 
.
Also, note that 
.
So, the inequality is really equal to 
.
We can then add 
to both sides to get .
So, the solution is , or in interval notation,
Answer:
Rectangle 1 area: 53 units squared
Shape 2 area: 43.75 units squared
Step-by-step explanation:
Shape 1:
We need to find the area of the two triangles in this rectangle and subtract it from the total area of the rectangle.
Total area of the rectangle:
13 * 5 = 65
Area of triangles (they are congruent):
1/2 * 4 * 3 = 6
6 * 2 = 12
65 - 12 = 53
The area of the shaded part is 53 (I don't know what your unit of measurement is. Just say units.)
Shape 2:
First, let's find the area of the parallelogram at the top.
10 * 3 = 30
Now, let's find the area of the rest of the shape. Triangle first...
1/2 * 4 * 1.5 = 3
Now, the rectangle and the tiny triangle we can section out at the end:
7 * 1.5 = 10.5 + 1/4 (Triangle) or 0.25 = 10.75
Total area: 10.75 + 3 + 30 = 43.75
The problem is asking us to isolate B. The given equation is solved for P, and we need to rearrange it for B.
First we need to square both sides. This will cancel out the square root on the right side.
P^2 = E + A^2/B^2
Next, subtract E from both sides.
P^2 - E = A^2/B^2
Next we need to get the B^2 out of the denominator. Multiply both sides by B^2.
B^2(P^2 - E) = A^2
Next divide both sides by (P^2 - E).
B^2 = A^2/(P^2 - E)
Lastly, take the square root of both sides.
B = sqrt(A^2/(P^2 - E))
