That's a question about the Pythagorean Theorem.
We use the Pythagorean Theorem to find the value of one side in a right triangle.
This theorem says that:

- <em>a</em> is the hypotenuse (It's the opposite side the right angle).
- <em>b</em> and <em>c </em>are cathetus (They're the adjacent sides the right angle).
Okay, now, let's go to solve this problem! In our figure, we have two cathetus. Their values are 10 and 7 and we have to find the value of <em>x </em>(the hypotenuse). Let's change this information in that formula.

Therefore, the value of <em>x </em>is
.
I hope I've helped. :D
Enjoy your studies! \o/
Answer:
The mean is the average of the numbers. It is easy to calculate: add up all the numbers, then divide by how many numbers there are. In other words it is the sum divided by the count.
The median is also the number that is halfway into the set. To find the median, the data should be arranged in order from least to greatest. If there is an even number of items in the data set, then the median is found by taking the mean (average) of the two middlemost numbers.
The range of a set of data is the difference between the highest and lowest values in the set. To find the range, first order the data from least to greatest. Then subtract the smallest value from the largest value in the set.
i hope i helped a bit
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
We have perpendicular bisector through a chord of the circle. We know the length, so either side of the chord is 11 due to the bisector cutting it directly in half. Since the radius is a fixed distance from the center to any point on the edge of the circle, we can draw the radius x from the circle to the end of the chord to form a right triangle.
We can use Pythagorean Theorem
to find the missing side length x. a=6, b=11 and c=x.


Answer:
i think the answer is A
Step-by-step explanation:
Because it says four times and 4 times 4 =16
sorry tried my best(don't write this part down)lol
We can work out the first three terms of this quadratic equation with substitution.
The first term of this sequence is where n=1, we can substitute 1 in for n
y=1^2+2(1)-4
y=1+2-4
y=-1
We can do this for the next two terms as well
y=2^2+2(2)-4
y=4+4-4
y=4
For the last term:
y=3^2+2(3)-4
y=9+6-4
y=11