There are 12 inches in a foot, so 9ft = 108in. Also, 80% = 0.8. Therefore the formula is:
h(n) = 108 * 0.8^n.
To find the bounce height after 10 bounces, substitute n=10 into the equation:
h(n) = 108 * 0.8^10 = 11.60in (2.d.p.).
Finally to find how many bounces happen before the height is less than one inch, substitute h(n) = 1, then rearrage with logarithms to solve for the power, x:
108 * 0.8^x = 1;
0.8^x = 1/108;
Ln(0.8^x) = ln(1/108);
xln(0.8) = ln(1\108);
x = ln(1/108) / ln(0.8) = -4.682 / -0.223 = 21 bounces
answer
6:15:16
step-by-step explanation
the ratio of stars to hearts to circles is stars:hearts:circles
plug in 6 for stars, 15 for hearts, and 16 for circles
6:15:16
since these do not have a common factor, this is your final answer
C + 2c + 12 = 75
Combine like terms:
3c + 12 = 75
Subtract 12 to both sides:
3c = 63
Divide 3 to both sides:
c = 21
Answer: (x-2)^2+(y+3)^2 = 9Side notes
1) This circle has a center of (2,-3)
2) The radius of this circle is 3
3) The graph is shown in the attached image
-------------------------------------------------------------------------------------
-------------------------------------------------------------------------------------
Work Shown:
x^2-4x+y^2+6y+4=0
x^2-4x+y^2+6y+4-4=0-4
x^2-4x+y^2+6y = -4
x^2-4x+4+y^2+6y = -4+4 ... see note 1 below
(x^2-4x+4)+y^2+6y = 0
(x-2)^2+y^2+6y = 0
(x-2)^2+y^2+6y+9 = 0+9 ... see note 2 below
(x-2)^2+(y^2+6y+9) = 9
(x-2)^2+(y+3)^2 = 9note 1: I'm adding 4 to both sides to complete the square for the x terms. You do this by first taking half of the x (not x^2) coefficient which in this case is -4. So take half of -4 to get -2. Then square this result to get 4
note 2: Like with note 1, I'm completing the square. What's different this time is that this is for the y terms now. The y coefficient is 6. Half of this is 3. Square 3 to get 9. So this is why we add 9 to both sides.
--------------------------------------------------------
So
the equation in standard form is (x-2)^2+(y+3)^2 = 9Note how
(x-2)^2+(y+3)^2 = 9
is equivalent to
(x-2)^2+(y-(-3))^2 = 3^2
So that second equation listed above is in the form (x-h)^2+(y-k)^2 = r^2
where
h = 2
k = -3
r = 3
making the center to be (h,k) = (2,-3) and the radius to be r = 3
The graph is attached.