B) the values will create a line with positive slope when graphed on a scatter plot.
Answer:
I think it's C) 18.5ft^2
it's just an educated guess though I haven't worked on area in a while
Answer:
okkk
Step-by-step explanation:
Check the picture below.
since the vertical distance, namely the y-coordinate, is twice as much as the horizontal, then if the horizontal is "x", the vertical one must be 2x.
let's find the hypotenuse first.
![\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{x}\\ b=\stackrel{opposite}{2x}\\ \end{cases} \\\\\\ c=\sqrt{x^2+(2x)^2}\implies c=\sqrt{x^2+4x^2}\implies c=\sqrt{5x^2}\implies c=x\sqrt{5} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3D%5Cstackrel%7Badjacent%7D%7Bx%7D%5C%5C%20b%3D%5Cstackrel%7Bopposite%7D%7B2x%7D%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7Bx%5E2%2B%282x%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7Bx%5E2%2B4x%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B5x%5E2%7D%5Cimplies%20c%3Dx%5Csqrt%7B5%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf sin(\theta )=\cfrac{\stackrel{opposite}{2~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }}{\stackrel{hypotenuse}{~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ \sqrt{5}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{\cfrac{2}{\sqrt{5}}\cdot \cfrac{\sqrt{5}}{\sqrt{5}}\implies \cfrac{2\sqrt{5}}{(\sqrt{5})^2}\implies \cfrac{2\sqrt{5}}{5}}](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B2~~%5Cbegin%7Bmatrix%7D%20x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B~~%5Cbegin%7Bmatrix%7D%20x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%5Csqrt%7B5%7D%7D%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Band%20rationalizing%20the%20denominator%7D~%5Chfill%20%7D%7B%5Ccfrac%7B2%7D%7B%5Csqrt%7B5%7D%7D%5Ccdot%20%5Ccfrac%7B%5Csqrt%7B5%7D%7D%7B%5Csqrt%7B5%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%7B5%7D%7D%7B%28%5Csqrt%7B5%7D%29%5E2%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%7B5%7D%7D%7B5%7D%7D)
Answer:
circle area = 1963.495 ft²
Step-by-step explanation:
<u>Given</u>
A rectangle 550 ft by 350 ft packed with circles on a rectangular grid
<u>Find</u>
the area of the largest circle that can be packed this way
<u>Solution</u>
The diameter of the circle must be a factor of both 350 and 550. We want the diameter to be the largest such factor. We can write the factors of 350 and 550 as ...
- 350 = 50 × 7
- 550 = 50 × 11
The largest factor common to both numbers is 50, so that will be the diameter of the largest circle that will fit on a square grid.
The area of a circle is commonly given by the formula ...
A = πr² . . . . . where r represents the radius of the circle.
The radius is half the diameter, so will be (50 ft)/2 = 25 ft. And the area of the circle is ...
A = π(25 ft)² = 625π ft² ≈ 1963.495 ft²
The area of one circle is about 1963.495 square feet.
_____
The total number of circles is 11×7 = 77. Their total area is 77×1963.495 ft² ≈ 151,189 ft². This is 151,189/192,400 ≈ 78.54% of the total area of the rectangle.
