We can use the Pythagorean Theorem to solve for side AB
a^2+b^2=c^2
a will be 6 and b will be 8 becuase those are the legs
6^2+8^2=c^2
36+64=c^2
100=c^2 (square root both sides)
c=10
Then we find the difference between 10 and 6 because 6 is the shortest leg
10-6=4 ft. so B
Hope this helps
These are so great! They are a perfect combination of Physics and pre-calculus! Your max height of that projectile is going to occur at the max value of the parabola, or at its vertex. So we need to find the vertex. The coordinates of the vertex will give us the x value, which is the time in seconds it takes to reach y which is the max height. Do this by completing the square. Begin by setting the equation equal to 0 and then moving the 80 over to the other side. Then factor out the -16. This is all that:

. Take half the linear term which is 4 and square it and add it in to both sides. Half of 4 is 2, 2 squared is 4, so add 4 into the set of parenthesis and to the -80.

. The -64 on the right comes from the fact that when you added 4 into the parenthesis, you had the -16 out in front which is a multiplier. -16 * 4 - -64. So what you really added in was -64. Now the perfect square binomial we created in that process was

. When we move the 144 back over by addition we find that the vertex of the polynomial is (2, 144). And that tells us that it takes 2 seconds for the projectile to reach its max height of 144 feet. To find the time interval in which the object's height decreases occurs from its max height of 144 to where the graph of the parabola goes through the x-axis to the right of the max. To find where the graph goes through the x-axis, or the zeroes of the graph, you factor the polynomial. When you do that using the quadratic formula you get that x = -1 and 5. So at its max height it is at 2 seconds, and by 5 seconds it hits the ground. So the time interval of its height decreasing is from 2 seconds to 5 seconds, or a total of 3 seconds. I think you need the 2 and 5, from the wording of your problem.
Answer:
32 1/5
Explanation:
Step 1 - Convert each mixed number into an improper fraction
4 1/5 • 7 2/3
21/5 • 23/3
Step 2 - Multiply straight across and simplify
21/5 • 23/3
483/15
161/5
32 1/5
Whoa just a second there.
What is the question ?
Is the question "Find the percent of increase ?"
In that case, the answer is 300% .
Here's why:
The amount of increase is 75.
The amount of increase as a fraction of
the original amount is
75 / 25 = 3.00 .
To change any number to a percent,
move the decimal point 2 places that way ===> .
3.00 = 300% .
______________________________
The question could have been different.
It could have been "What is the amount of the increase ?"
That's 75.
It could have been "How many times larger is 100 than 25 ?"
100 is 4 times as large as 25, and 3 times larger.
It could have been "How many times has 25 been increased ?"
This is a terrible question ... there are a lot of those in schools.
The answer could be "25 was only increased once, by 75."
The answer could be "25 was increased to 4 times its size."
The answer could be "25 was increased by 3 times its size."
Answer:
The 95% confidence interval for the difference in mean profile height and mean actual height of online daters is between 0.27 and 0.87 inches.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 30 - 1 = 29
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 29 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.0452
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 0.57 - 0.3 = 0.27 inches.
The upper end of the interval is the sample mean added to M. So it is 0.57 + 0.3 = 0.87 inches.
The 95% confidence interval for the difference in mean profile height and mean actual height of online daters is between 0.27 and 0.87 inches.