He will need 15 loads to cover his entire driveway.
1/5 of the driveway would be 3 loads
3x5=15
Answer:
15^3−31^2+20−4
Step-by-step explanation:
Distribute
(3−2)(5^2−7+2)
3(5^2−7+2)−2(5^2−7+2)
Distribute
3(52−7+2)−2(5^2−7+2)
15^3−21^2+6−2(5^2−7x+2)
Distribute
15^3−21^2+6−2(5^2−7+2)
15^3−21^2+6−10^2+14−4
Combine like terms
15^3−21^2+6−10^2+14−4
15^3−31^2+6+14−4
Combine like terms
15^3−31^2+6+14−4
15^3−31^2+20−4
Solution
15^3−31^2+20−4
![\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{1})~\hspace{10em} slope = m\implies -6 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-1=-6(x-3) \\\\\\ y-1=-6x+18\implies y=-6x+19](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B3%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29~%5Chspace%7B10em%7D%20slope%20%3D%20m%5Cimplies%20-6%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-1%3D-6%28x-3%29%20%5C%5C%5C%5C%5C%5C%20y-1%3D-6x%2B18%5Cimplies%20y%3D-6x%2B19)
bearing in mind that
standard form for a linear equation means
• all coefficients must be integers, no fractions
• only the constant on the right-hand-side
• all variables on the left-hand-side, sorted
• "x" must not have a negative coefficient
then we'd end up with 6x + y = 19.
The perpendicular distance from Frank to the height of statue of Liberty is 306-5.5 = 300.5 ft .
And frank stand 400 feet away from the base of the statue , so base=400 feet
Since tangent function establish relationship between height and base, so we use tanhere.
![tan \ x = \frac{300.5}{400} => x = 36.9 degree](https://tex.z-dn.net/?f=%20tan%20%5C%20x%20%3D%20%5Cfrac%7B300.5%7D%7B400%7D%20%3D%3E%20x%20%3D%2036.9%20degree%20%20)
So the correct option is the first option .
Answer:
115m
Step-by-step explanation:
Note that one of the face of a pyramid is a right angled triangle.
Since the pyramid was about 147m tall, the height of triangle will be 147m. If one of the faces is built with one of its faces at 52° to the incline, we can find the original length of one of its sides using SOH CAH TOA
tan(theta) = opposite/adjacent
Given theta = 52°, opposite = 147m (the side directly opposite the angle)
Substituting we have;
Tan52° = 147/adjacent
Adjacent = 147/tan52°
Adjacent = 114.8m
= 115m
Therefore 115m is the original length of one of its side