Answer:
![67.5\text{ [square units]}](https://tex.z-dn.net/?f=67.5%5Ctext%7B%20%5Bsquare%20units%5D%7D)
Step-by-step explanation:
The composite figure consists of one rectangle and two triangles. We can add up the area of these individual shapes to find the total area of the irregular figure.
<u>Formulas</u>:
- Area of rectangle with base
and height 
: 
- Area of triangle with base and height :
By definition, the base and height must intersect at a 90 degree angle.
The rectangle has a base of 10 and a height of 5. Therefore, its area is 
.
The smaller triangle to the left of the rectangle has a base of 2 and a height of 5. Therefore, its area is 
.
Finally, the larger triangle on top of the rectangle has a base of 5 and a height of 5. Therefore, its area is 
.
Thus, the area of the total irregular figure is:
![50+5+12.5=\boxed{67.5\text{ [square units]}}](https://tex.z-dn.net/?f=50%2B5%2B12.5%3D%5Cboxed%7B67.5%5Ctext%7B%20%5Bsquare%20units%5D%7D%7D)
Answer:
246 ft is the maximum height
Step-by-step explanation:
The height h given above is a quadratic function. The graph of h as a function of time t gives a parabolic shape and the maximum height h occur at the vertex of the parabola. For a quadratic function of the form h = a t² + bt + c, the vertex is located at t = - b / 2a. Hence for h given above the vertex in the question s(t) = 124 + 64t − 16t², is at t
t = -64/2(-16) = 64/32 = 2 seconds
Thus, 2 seconds after the object was thrown, it reaches its highest point (maximum value of h) which is given by
h = -16(2)² + 64 (2) + 124 = 246eet
Pi is multiplied by diameter so,
35 x pi= 109.995
tell me how many decimal places should the finals should be so that I can round off
Second option is the correct one.
-24, -96, -384, -1536, -6144, -24576
Answer:
y = (2/5) OR y = (6/5)
Step-by-step explanation:
The first step is isolating the expression within the absolute value bars. The first thing we can do is subtract both sides by 8. If we do that, we get -2|4-5y| = -4. Now, to completely isolate the absolute value, we would have to divide by -2. This yields |4 - 5y| = 2. Finally, we can remove the absolute value bars. However, to do this, we need to first understand what an absolute value bar does to an equation. Lets say that |x| = 2. Absolute value describes the DISTANCE of some quantity from 0 (on the number line). Therefore, x (which is inside the absolute value bars) can be either positive or negative 2 (they are BOTH two units away from 0). Similarly, in this case, (4 - 5y) can either be 2 or -2 (because the absolute value of both is 2). Now we have two possible solutions to solve for:
4 - 5y = 2 OR 4 - 5y = -2
5y = 2 OR 5y = 6
y = (2/5) OR y = (6/5)
If we plug both of these answers back into the equation we can see that they both check out.
