The equation gives the height of the ball. That is, h is the height of the ball. t is the time. Since we are looking for the time at which the height is 8 (h=8), we need to set the equation equal to 8 and solve for t. We do this as follows:




This is a quadratic equation and as it is set equal to 0 we can solve it using the quadratic formula. That formula is:

You might recall seeing this as "x=..." but since our equation is in terms of t we use "t-=..."
In order to use the formula we need to identify a, b and c.
a = the coefficient (number in front of)

= 16.
b = the coefficient of t = -60
c = the constant (the number that is by itself) = 7
Substituting these into the quadratic formula gives us:



As we have "plus minus" (this is usually written in symbols with a plus sign over a minus sign) we split the equation in two and obtain:

and

So the height is 8 feet at t = 3.63 and t=.12
It should make sense that there are two times. The ball goes up, reaches it's highest height and then comes back down. As such the height will be 8 at some point on the way up and also at some point on the way down.
The equation which is equivalent to
is
or x = 6 (
).
<u>Step-by-step explanation:</u>
Given Equation:

As we know, in terms of logarithmic rules, when b is raised to the power of y is equal x:

Then, the base b logarithm of x is equal to y

Now, use the logarithmic rule for the given equation by comparing with above equation. We get b = x, y = 2, and x = 36. Apply this in equation,


When taking out the squares on both sides, we get x = 6. Hence, the given equation can be written as 
Answer:
378$
Step-by-step explanation:
since u got 700ft^2 you would multiply 700 with 0.54 and get 378$
Answer:
13%
15%
41%
31%
Step-by-step explanation:
To find the percentage one amount is out of another amount, divide the smaller amount by the bigger amount. You will end up with a decimal (0.xy where x is the 10% place and y is the 1% place).
For example: to find the percentage that 39 is out of 300 students, we divide 39 by 300. This gives us the decimal 0.13. Written as a percentage, that is 13%.
We can do this for all of them.
Here is the math for each one:
39 ÷ 300 = 0.13 → 13%
45 ÷ 300 = 0.15 → 15%
123 ÷ 300 = 0.41 → 41%
93 ÷ 300 = 0.31 → 31%
Once you have all these decimals, you might choose to check your work by adding them up. They should at up to 1.00, or 100%. In this case, they do, so we know these answers are correct.
I hope this helps. Please let me know if you have questions.
Answer:
bxbdbdbdbhd
Step-by-step explanation:
$2200jfjfjffjfodod/month