Answer:
m= 13/7
Step-by-step explanation:
(-5, 2) , (-12, -11)
Slope(m)
m= (y2-y1)/(x2-x1)
m= (-11 - 2)/(-12 + 5)
m= (-13)/(-7)
m= 13/7
The computation shows that the placw on the hill where the cannonball land is 3.75m.
<h3>How to illustrate the information?</h3>
To find where on the hill the cannonball lands
So 0.15x = 2 + 0.12x - 0.002x²
Taking the LHS expression to the right and rearranging we have:
-0.002x² + 0.12x -.0.15x + 2 = 0.
So we have -0.002x²- 0.03x + 2 = 0
I'll multiply through by -1 so we have
0.002x² + 0.03x -2 = 0.
This is a quadratic equation with two solutions x1 = 25 and x2 = -40 since x cannot be negative x = 25.
The second solution y = 0.15 * 25 = 3.75
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Complete question:
The flight of a cannonball toward a hill is described by the parabola y = 2 + 0.12x - 0.002x 2 . the hill slopes upward along a path given by y = 0.15x. where on the hill does the cannonball land?
Answer: 3/10
Step-by-step explanation:
3/10 + 4/10= 7/10
10/10 - 7/10= 3/10
The volume of the solid that lies inside both the cylinder x² + y² = 1 and the sphere x² + y² + z² = 16 is 24.74 cubic units.
<h3>What is the volume of the solid?</h3>
Find the volume of the solid that lies inside both the cylinder x² + y² = 1 and the sphere x² + y² + z² = 16
From the sphere and cylinder, the cylindrical coordinate will be
And the radius of the cylinder is |r| < 1 and the 0 ≤ θ ≤ 2π.
Then the volume will be given as
![\rm V = \int _0^{2\pi} \int _0^1 \int _{- \sqrt{16 - r^2}}^{\sqrt{16-r^2}} \left ( r \ dz \ dr \ d\theta \right ) \\\\\\V = 2\pi \int_0^1 \left ( 2r\sqrt{16-r^2} \right ) \ dr\\\\\\V = 4\pi \left [ -\dfrac{1}{3} (16 - r^2)^{3/2} \right ]\\\\\\V = \dfrac{4\pi}{3} \left ( 16^{3/2} - 15^{3/2} \right )\\\\\\V = 24.74](https://tex.z-dn.net/?f=%5Crm%20V%20%3D%20%5Cint%20_0%5E%7B2%5Cpi%7D%20%5Cint%20_0%5E1%20%5Cint%20_%7B-%20%5Csqrt%7B16%20-%20r%5E2%7D%7D%5E%7B%5Csqrt%7B16-r%5E2%7D%7D%20%5Cleft%20%28%20r%20%5C%20dz%20%5C%20dr%20%5C%20d%5Ctheta%20%20%5Cright%20%29%20%5C%5C%5C%5C%5C%5CV%20%3D%202%5Cpi%20%5Cint_0%5E1%20%5Cleft%20%28%202r%5Csqrt%7B16-r%5E2%7D%20%5Cright%20%29%20%5C%20dr%5C%5C%5C%5C%5C%5CV%20%3D%204%5Cpi%20%5Cleft%20%5B%20-%5Cdfrac%7B1%7D%7B3%7D%20%2816%20-%20r%5E2%29%5E%7B3%2F2%7D%20%5Cright%20%5D%5C%5C%5C%5C%5C%5CV%20%3D%20%5Cdfrac%7B4%5Cpi%7D%7B3%7D%20%5Cleft%20%28%2016%5E%7B3%2F2%7D%20-%2015%5E%7B3%2F2%7D%20%5Cright%20%29%5C%5C%5C%5C%5C%5CV%20%3D%2024.74)
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