Answer:
I think the answer is B 23,345
Answer:
The equation of the line with slope 6 and containing the point (0, 4) will be:
![y = 6x+4](https://tex.z-dn.net/?f=y%20%3D%206x%2B4)
The graph of the line y = 6x+4 is attached below.
Step-by-step explanation:
The slope-intercept form of the line equation
![y = mx+b](https://tex.z-dn.net/?f=y%20%3D%20mx%2Bb)
where
Given
The y-intercept can be determined by setting x = 0 and determining the corresponding value of y.
The point (0, 4) indicates that:
at x = 0, y = 4
Thus, the y-intercept b = 4
now substituting b = 4 and m = 6 in the slope-intercept form of the line equation
![y = mx+b](https://tex.z-dn.net/?f=y%20%3D%20mx%2Bb)
![y = 6x+4](https://tex.z-dn.net/?f=y%20%3D%206x%2B4)
Thus, the equation of the line with slope 6 and containing the point (0, 4) will be:
![y = 6x+4](https://tex.z-dn.net/?f=y%20%3D%206x%2B4)
The graph of the line y = 6x+4 is attached below.
Answer:
Water needed for pool = 486 cubic feet
Plastic liner required for the pool = 298.3 feet
Step-by-step explanation:
Top view of the pool is a composite figure, having one rectangle and a trapezoid.
1). Water needed for the pool = volume of the pool
Volume of the pool = Area of the base × Depth
= (Area of the rectangle + Area of the trapezoid)× depth
Area of the trapezoid = ![\frac{1}{2}(b_{1}+b_{2})\times h](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28b_%7B1%7D%2Bb_%7B2%7D%29%5Ctimes%20h)
= ![\frac{1}{2}(11+9)\times (12-1.5)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%2811%2B9%29%5Ctimes%20%2812-1.5%29)
= 105 ft²
Area of the rectangle = Length × width
= 11 × 1.5
= 16.5 ft²
Now, volume of the pool = (105 + 16.5) × 4
= 121.5 × 4
= 486 cubic feet
b). Liner required = surface area of the pool excluding top
= Surface area of the walls + Area of the pool base
= (Perimeter of the pool) × depth + area of the base
= (12 + 11 + 1.5 + 10.7 + 9)×4 + 121.5
= 176.8 + 121.5
= 298.3 square feet
Therefore, amount of water required = 486 cubic feet
liner needed = 298.3 square feet
First lets see the pythagorean identities
![sin^2 \Theta + cos^2 \Theta =1](https://tex.z-dn.net/?f=%20sin%5E2%20%5CTheta%20%2B%20cos%5E2%20%5CTheta%20%3D1%20)
So if we have to solve for sin theta , first we move cos theta to left side and then take square root to both sides, that is
![sin^2 \Theta = 1-cos^2 \Theta => sin \theta = \pm \sqrt{1-cos^2 \Theta}](https://tex.z-dn.net/?f=%20sin%5E2%20%5CTheta%20%3D%201-cos%5E2%20%5CTheta%20%3D%3E%20sin%20%5Ctheta%20%3D%20%5Cpm%20%5Csqrt%7B1-cos%5E2%20%5CTheta%7D%20)
Now we need to check the sign of sin theta
First we have to remember the sign of sin, cos , tan in the quadrants. In first quadrant , all are positive. In second quadrant, only sin and cosine are positive. In third quadrant , only tan and cot are positive and in the last quadrant , only cos and sec are positive.
So if theta is in second quadrant, then we have to positive sign but if theta is in third or fourth quadrant, then we have to use negative sign .