X²/9 + y²/16 = 1
The general for mula of an ellipse is:
(x-h)²/a² + (y-k)²/b² = 1. Since h = k = 0, this ellipse
x²/9 + y²/16 = 1
passes by the center O. (Note that a = 3 and b = 4)
Moreover sin b>a so the majore axis is (on the y-axis) and a the minor axis (on the x-axis) [Its shape is as a vertical egg].
On a system of perpendicular axis that intercept in O., take on the x-axis 2 points A & B with respective coordinate A(3.0) and B(-3,0)
On the y-axis your report A'(0,+4) and B'(0,-4). So you have the vertex of the ellipse, then it's easy to draw it
Pythagorean Theorem is:
c² = a² + b²
Let’s find the values of a and b.
Looking at the image provided in the question we have 2 number already, 70 and 120.
Let’s name the side with length 120 as a, and the side with length 70 as b.
Then we should substitute these values into the Pythagorean Theorem:
c² = 120² + 70²
c² = 14400 + 4900
c² = 19300
Then we should work out the value of c by square rooting both sides:
√c = √19300
c = 138.92444
Therefore the ball would have to be hit a total of 138.92444 units.
If you have to round, that would be 139 units.
Answer:
the surface area of the decoration is 31.5
Step-by-step explanation:
So to solve you need to set up equations, Using h as the height. So the base is 9 inches more (+9) than 3 times the height (3h) so 3h+9 equals the base. You have the area so you need to plug in the equation for the base and h for height and divide it all by 2. h(3h+9)/2=105. after you solve that and get h by itself you should get h= 7 and the b= 30
