Answer:
153
Step-by-step explanation:
Answer:
35
Step-by-step explanation:
We use trignometry here
Assuming that the angle created between the surface of the sea and the 56feet below line is 90 degrees.
So we use the sin rule
sin90/97 = sin x/56
multiply 56 to the other side
56*sin90/97 = sin x
sin90=1
56/97 = sin x
x= sin^-1 (56/97)
x= 35.3
Round and you get 35
The way I'm reading this, your equation would be 5+0.75x=y.
If 2, 5, 8, and 10 are units, then you would plug those in individually and solve, like this:
5+0.75(2)= 6.5
5+0.75(5)=8.75
5+0.75(8)=11
5+0.75(10)=12.5
Hope this helps!
Answer:
2 feet
Step-by-step explanation:
Let 'w' be twice the width of the deck.
The area of the swimming pool is:
![A_p = \frac{\pi d^2}{4}\\A_p = \frac{\pi 28^2}{4} = 196\pi\\](https://tex.z-dn.net/?f=A_p%20%3D%20%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7D%5C%5CA_p%20%3D%20%5Cfrac%7B%5Cpi%2028%5E2%7D%7B4%7D%20%3D%20196%5Cpi%5C%5C)
The area of the swimming pool plus the deck is:
![A_{p+d} = \frac{\pi (d+w)^2}{4}\\](https://tex.z-dn.net/?f=A_%7Bp%2Bd%7D%20%3D%20%5Cfrac%7B%5Cpi%20%28d%2Bw%29%5E2%7D%7B4%7D%5C%5C)
Therefore, the area of the deck is given by:
![A_d = A_{p+d} - A_p\\A_d = 60\pi = \frac{\pi (28+w)^2}{4} - 196\pi\\1,024 = (28+w)^2\\w^2+56w - 240 = 0](https://tex.z-dn.net/?f=A_d%20%3D%20A_%7Bp%2Bd%7D%20-%20A_p%5C%5CA_d%20%3D%2060%5Cpi%20%3D%20%5Cfrac%7B%5Cpi%20%2828%2Bw%29%5E2%7D%7B4%7D%20-%20196%5Cpi%5C%5C1%2C024%20%3D%20%2828%2Bw%29%5E2%5C%5Cw%5E2%2B56w%20-%20240%20%3D%200)
Solving the quadratic equation:
![w^2+56w - 240 = 0\\w= -56 \pm\frac{\sqrt{56^2-(4*1*-240)} }{2} \\w_1 = -60\\w_2 = 4](https://tex.z-dn.net/?f=w%5E2%2B56w%20-%20240%20%3D%200%5C%5Cw%3D%20-56%20%5Cpm%5Cfrac%7B%5Csqrt%7B56%5E2-%284%2A1%2A-240%29%7D%20%7D%7B2%7D%20%5C%5Cw_1%20%3D%20-60%5C%5Cw_2%20%3D%204)
Since the width cannot be negative, w = 4 feet, ad the width is 2 feet.
3n-1
where n = 1 ; n = 2 ; n = 3
3(n) - 1 = 3(1) - 1 = 3 - 1 = 2
3(n) - 1 = 3(2) - 1 = 6 - 1 = 5
3(n) - 1 = 3(3) - 1 = 9 - 1 = 8