Answer:
Step-by-step explanation:
Perpendicular cross section of a pyramid ------- triangle.
Imagine you cut pyramid like that. Height of pyramid would be height of your triangle. and sides of pyramid would be sides of triangle.
Perpendicular cross section of a cylinder ------- rectangle
diameter of a base of cylinder would be one side of rectangle and height of cylinder would be other side of rectangle. (pair of sides)
parallel cross section of a sphere ------ circle
when you cut sphere a side would appear with shape of circle.
shape create.... ----- cylinder.
one side of rectangle would be high of cylinder and other would be radius of it.
last one is cone. If you place right angle in coordinate beginning (0.0) and one of shorter sides to lie on y axis other would be on x axis. when rotated you get cone.
Answer:
10 pounds of trial mix that costs $2.45 and 20 pounds of trial mix that costs $2.30
Step-by-step explanation:
![\\\text{Let x pounds of the trial mix that costs \$2.45 per pound, and y pounds of the trial mix}\\\text{that costs \$2.30 per pound are added to get a 30 pounds of trial mix }\\\text{that costs \$2.35 per pound. since the total mixture is 30 pounds, so we get}\\\\x+y=30\ \ \ \ \ \ \ \ \ \ \ \ \ ...... (i)\\\\\text{and the total cost equation is given by}\\\\\$2.45 x + \$2.30 y=\$2.35 (30)\\\\2.45 x + 2.30 y=70.5 \ \ \ \ \ \ \ \ \ \ \ \ \ ...... (ii)\\](https://tex.z-dn.net/?f=%5C%5C%5Ctext%7BLet%20x%20pounds%20of%20the%20trial%20mix%20that%20costs%20%5C%242.45%20per%20pound%2C%20and%20y%20pounds%20of%20the%20trial%20mix%7D%5C%5C%5Ctext%7Bthat%20costs%20%5C%242.30%20per%20pound%20are%20added%20to%20get%20a%2030%20pounds%20of%20trial%20mix%20%7D%5C%5C%5Ctext%7Bthat%20costs%20%5C%242.35%20per%20pound.%20since%20the%20total%20mixture%20is%2030%20pounds%2C%20so%20we%20get%7D%5C%5C%5C%5Cx%2By%3D30%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%20......%20%28i%29%5C%5C%5C%5C%5Ctext%7Band%20the%20total%20cost%20equation%20is%20given%20by%7D%5C%5C%5C%5C%5C%242.45%20x%20%2B%20%5C%242.30%20y%3D%5C%242.35%20%2830%29%5C%5C%5C%5C2.45%20x%20%2B%202.30%20y%3D70.5%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20......%20%28ii%29%5C%5C)
![\text{now to solve the equation (i) and (ii), we'll substitute }y=30-x \text{ from (i) into (ii)}\\\text{so we get}\\\\2.45x+2.30(30-x)=70.5\\\\2.45x+69-2.30x=70.5\\\\\Rightarrow 2.45x-2.30x=70.5-69\\\\\Rightarrow 0.15x=1.5\\\\\Rightarrow x=\frac{1.5}{0.15}\\\\\Rightarrow x=10\\\\\text{plug this value of x in (i), we get}\\\\y=30-x=30-10=20\\\\\text{so we must add 10 Pounds of trial mix that costs \$ 2.45 and 20 pounds of}\\\text{the trial mix that costs \$2.30}](https://tex.z-dn.net/?f=%5Ctext%7Bnow%20to%20solve%20the%20equation%20%28i%29%20and%20%28ii%29%2C%20we%27ll%20substitute%20%7Dy%3D30-x%20%5Ctext%7B%20from%20%28i%29%20into%20%28ii%29%7D%5C%5C%5Ctext%7Bso%20we%20get%7D%5C%5C%5C%5C2.45x%2B2.30%2830-x%29%3D70.5%5C%5C%5C%5C2.45x%2B69-2.30x%3D70.5%5C%5C%5C%5C%5CRightarrow%202.45x-2.30x%3D70.5-69%5C%5C%5C%5C%5CRightarrow%200.15x%3D1.5%5C%5C%5C%5C%5CRightarrow%20x%3D%5Cfrac%7B1.5%7D%7B0.15%7D%5C%5C%5C%5C%5CRightarrow%20x%3D10%5C%5C%5C%5C%5Ctext%7Bplug%20this%20value%20of%20x%20in%20%28i%29%2C%20we%20get%7D%5C%5C%5C%5Cy%3D30-x%3D30-10%3D20%5C%5C%5C%5C%5Ctext%7Bso%20we%20must%20add%2010%20Pounds%20of%20trial%20mix%20that%20costs%20%5C%24%202.45%20and%2020%20pounds%20of%7D%5C%5C%5Ctext%7Bthe%20trial%20mix%20that%20costs%20%5C%242.30%7D)
86/139*100%= 61.87%
The final answer is 61.87%~
Answer:
(Including the units:
)
Step-by-step explanation:
The missing figure is attached.
1. Find the area of the semi-circle with this formula:
![A=\frac{\pi r^2}{2}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%20r%5E2%7D%7B2%7D)
Where "r" is the radius.
You can observe in the figure that the diameter of the semi-circle is:
![d=15\ ft](https://tex.z-dn.net/?f=d%3D15%5C%20ft)
Since the radius is half the diameter:
![r=\frac{15\ ft}{2}=7.5\ ft](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B15%5C%20ft%7D%7B2%7D%3D7.5%5C%20ft)
Substituting the radius into the formula, you get:
![A_1=\frac{\pi (7.5\ ft)^2}{2}=88.35\ ft^2](https://tex.z-dn.net/?f=A_1%3D%5Cfrac%7B%5Cpi%20%287.5%5C%20ft%29%5E2%7D%7B2%7D%3D88.35%5C%20ft%5E2)
2. The area of a rectangle can be calculated with this formula:
![A=lw](https://tex.z-dn.net/?f=A%3Dlw)
Where "l" is the length and "w" is the width.
Since the tile won't be put in the island located at kitchen, you need to subtract the area of the smaller rectangle from the area of the larger rectangle. Then:
Therefore, the total area is:
![A_t=88.35\ ft^2+650\ ft^2=738.35](https://tex.z-dn.net/?f=A_t%3D88.35%5C%20ft%5E2%2B650%5C%20ft%5E2%3D738.35)
Since tile can only be sold in multiples of 10 square feet, then he needs to buy ![800\ ft^2](https://tex.z-dn.net/?f=800%5C%20ft%5E2)