Answer:
87.5 square inches
Step-by-step explanation:
The belt knuckle as shown in the diagram above consist of a 4 rectangles (2 are the of the same dimension, the other two are of the same direction differently) and a rectangle.
Area of the belt knuckle = area of the 4 triangles + area of the rectangle
✔️Area of the 2 rectangles with the following dimensions:
base (b) = 9 inches
height (h) = 2.5 inches
Area of the two triangles = 2(½*b*h)
= 2(½*9*2.5)
= 22.5 inches²
✔️Area of the 2 rectangles with the following dimensions:
base (b) = 4 inches
height (h) = 5 inches
Area of the two triangles = 2(½*b*h)
= 2(½*4*5)
= 20 inches²
✔️Area of the rectangle = l*w
l = 9 inches
w = 5 inches
Area = 9*5 = 45 inches²
✔️Area of the belt knuckle = 22.5 + 20 + 45 = 87.5 square inches
9514 1404 393
Answer:
Step-by-step explanation:
Add 9 to both sides of the equation, and divide by the coefficient of B.
T = BA -9
T +9 = BA
(T +9)/A = B
We can rewrite that with B on the left:
B = (T +9)/A
Answer:
$49.60
Step-by-step explanation:
The equation setup I used for this was as follows: $16 x (112/36)
Using PEMDAS, you'd start with the division part, 112/36. Although, I did this a bit differently, and being a bit further in school, I don't know if you go through this where you're at.
Start by factoring 112 into 4(28). This leaves you with 4(28)/36.
Next, cancel 4 out of 36. Since 4 is the common factor in 28 and 36, you cancel the four out of the equation, leaving you with 28/9. Convert this to its decimal form, 3.1.
Finally, take the 3.1 and multiply that by $16, which comes out to $49.60.
C I think not sure I think tho.
Answer:
3π square units.
Step-by-step explanation:
We can use the disk method.
Since we are revolving around AB, we have a vertical axis of revolution.
So, our representative rectangle will be horizontal.
R₁ is bounded by y = 9x.
So, x = y/9.
Our radius since our axis is AB will be 1 - x or 1 - y/9.
And we are integrating from y = 0 to y = 9.
By the disk method (for a vertical axis of revolution):
![\displaystyle V=\pi \int_a^b [R(y)]^2\, dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%20V%3D%5Cpi%20%5Cint_a%5Eb%20%5BR%28y%29%5D%5E2%5C%2C%20dy)
So:
Simplify:
Integrate:
![\displaystyle V=\pi\Big[y-\frac{1}{9}y^2+\frac{1}{243}y^3\Big|_0^9\Big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20V%3D%5Cpi%5CBig%5By-%5Cfrac%7B1%7D%7B9%7Dy%5E2%2B%5Cfrac%7B1%7D%7B243%7Dy%5E3%5CBig%7C_0%5E9%5CBig%5D)
Evaluate (I ignored the 0):
![\displaystyle V=\pi[9-\frac{1}{9}(9)^2+\frac{1}{243}(9^3)]=3\pi](https://tex.z-dn.net/?f=%5Cdisplaystyle%20V%3D%5Cpi%5B9-%5Cfrac%7B1%7D%7B9%7D%289%29%5E2%2B%5Cfrac%7B1%7D%7B243%7D%289%5E3%29%5D%3D3%5Cpi)
The volume of the solid is 3π square units.
Note:
You can do this without calculus. Notice that R₁ revolved around AB is simply a right cone with radius 1 and height 9. Then by the volume for a cone formula:
We acquire the exact same answer.
