Answer:
x = 12
Step-by-step explanation:
The segment from the vertex to the base is a perpendicular bisector.
Using Pythagoras' identity on the right triangle on the right.
(
x )² + 3² = (
)² , that is
x² + 9 = 45 ( subtract 9 from both sides )
x² = 36 ( multiply both sides by 4 )
x² = 144 ( take the square root of both sides )
x = 
= 12
Answer:
117.5 ft²
Step-by-step explanation:
To find the area of the shaded area, we can find the area of the square, then subtract the areas of the two semicircles.
First, we will find the area of the given square, by using the formula 
, multiplying the length and width. The dimensions of this square are 14×14.
14 · 14 = 196
The area of the square is 196 ft².
We can now find the area of the two congruent half-circles. Since they are identical, we can simply find the area of one circle if it was whole. To find the area of a circle, we'll use the formula 
. With some simple deduction, we can see that the diameter of the circle is 10 ft, so the radius would be 5 ft long. Plug our values into the formula.
A = 
5²
We will use 3.14 for .
A = 78.5
The area of both the semicircles is 78.5 ft².
Now, we can subtract.
196 - 78.5 = 117.5
The area of the figure is 117.5 ft².
Good luck ^^
.32 multiplied by 300 is 96
Find the mean, median, and mode of the data set. Round to the nearest tenth. 15, 13, 9, 9, 7, 1, 11, 10, 13, 1, 13 mean = 8.5, m
jasenka [17]
Answer:
Mean = 9.3
Median = 10
Mode = 13
Step-by-step explanation:
To get the mean, you would have to add up all of the numbers (102), then divide how many numbers you have with the numbers added up (9.3). To get the median, you would have to put all of the numbers that you have, in order, from least to greatest and cross off each number, one from each side before getting to the number in the middle (unless if you have an even number; works better with odd numbers, which is easier). To get the mode, you would need to find out which number appears the most, and, if there is more than one, you can put the numbers down. Hope this helps.
First list all the terms out.
e^ix = 1 + ix/1! + (ix)^2/2! + (ix)^3/3! ...
Then, we can expand them.
e^ix = 1 + ix/1! + i^2x^2/2! + i^3x^3/3!...
Then, we can use the rules of raising i to a power.
e^ix = 1 + ix - x^2/2! - ix^3/3!...
Then, we can sort all the real and imaginary terms.
e^ix = (1 - x^2/2!...) + i(x - x^3/3!...)
We can simplify this.
e^ix = cos x + i sin x
This is Euler's Formula.
What happens if we put in pi?
x = pi
e^i*pi = cos(pi) + i sin(pi)
cos(pi) = -1
i sin(pi) = 0
e^i*pi = -1 OR e^i*pi + 1 = 0
That is Euler's identity.
