Answer:
C) √221 units
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra II</u>
- Distance Formula:
![d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
Step-by-step explanation:
<u>Step 1: Define</u>
Point (-7, 7)
Point (7, 2)
<u>Step 2: Find distance </u><em><u>d</u></em>
Simply plug in the 2 coordinates into the distance formula to find distance <em>d</em>
- Substitute [DF]:
![d = \sqrt{(7+7)^2+(2-7)^2}](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7B%287%2B7%29%5E2%2B%282-7%29%5E2%7D)
- Add/Subtract:
![d = \sqrt{(14)^2+(-5)^2}](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7B%2814%29%5E2%2B%28-5%29%5E2%7D)
- Exponents:
![d = \sqrt{196+25}](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7B196%2B25%7D)
- Add:
![d = \sqrt{221}](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7B221%7D)
Answer:
<u>2 </u>sack of cement is required to cover Joel's entire floor.
Step-by-step explanation:
Given:
A sack cement can cover area = 45 sq. ft
Length of Joel floor = 8 ft
Width of Joel Floor = 7 ft
We will first find the Area of Joel's Floor
Area is the Product of length and width.
Area of the Floor = ![length \times width](https://tex.z-dn.net/?f=length%20%5Ctimes%20width)
Substituting the values we get;
Area of the Floor = ![8\times 7 = 56\ ft^2](https://tex.z-dn.net/?f=8%5Ctimes%207%20%3D%2056%5C%20ft%5E2)
Now Given:
1 sack of cement cover area = 45 sq ft
Number of sack cement required to cover area = 56 sq. ft.
By Using Unitary method we will find the same.
∴ Number of sack cement required = ![\frac{56}{45} = 1.24](https://tex.z-dn.net/?f=%5Cfrac%7B56%7D%7B45%7D%20%3D%201.24)
Since Sack of cement cannot be in point value hence we can say 2 sack cement is required.
Hence we can say 2 sack of cement is required to cover Joel's entire floor.
Answer:
56844.9 units squared
Step-by-step explanation:
The surface area of a cone is denoted by:
, where r is the radius and l is the slant height. The slant height is basically the length from a point on the base circle to the top vertex of the cone.
Here, since our diameter is 50 and diameter is twice the radius, then our radius is r = 50/2 = 25.
To find the slant height, we have to use the Pythagorean Theorem:
, where h is the height
![l^2=25^2+50^2=625+2500=3125](https://tex.z-dn.net/?f=l%5E2%3D25%5E2%2B50%5E2%3D625%2B2500%3D3125)
![\sqrt{l^2} =\sqrt{3125}](https://tex.z-dn.net/?f=%5Csqrt%7Bl%5E2%7D%20%3D%5Csqrt%7B3125%7D)
![l=25\sqrt{5}](https://tex.z-dn.net/?f=l%3D25%5Csqrt%7B5%7D)
Now, plug these values of r and l into the first equation above:
![A=\pi r^2+\frac{1}{2} \pi r^2*l](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cpi%20r%5E2%2Al)
≈ 56844.9 units squared
Answer:
0
Step-by-step explanation:
Here, given the function f(x), we want to calculate f(1)
Mathematically, this means we shall substitute 1 for the value of x in that function.
Thus, given F(x) = 1-x
Then F(1) would be = 1-1 = 0
Pienso que un libro pesa 1.9 libras