82
*41
2*1=2
8*1=8
4*2=8
8*4=32
328+82=410
Answer:
<em>AXE</em> and <em>CXD </em>are vertical angles
<em>AXF</em> and <em>FXD </em>are supplementary angles
<em>DXC </em>and <em>BXC </em>are complementary angles
<em>EXA</em> and <em>AXB </em>are adjacent angles
<em>AXC </em>and <em>CXD</em> are supplementary angles
<em>EXD </em>and <em>AXC </em>are vertical angles
Step-by-step explanation:
1. a. Pretty much, you just have to rearrange it so that the highest power is in the front. So, here's your answer:
b. It's a 4th-degree polynomial. A degree means that "what's the highest power?"
c. It's a trinomial. It has 3 terms, hence it's a
trinomial.
2. a. Since it's an odd power and a negative coefficient, it will be:
x→∞, f(x)→-∞
x→-∞, f(x)→∞
b. The degree is even and the coefficient is negative, so it will be:
x→∞, f(x)→-∞
x→-∞, f(x)→-∞
3. a. This basically means that if you solve for x, you should get -2, 1, and 2. So, to do this, you can just write it in factored form and multiply inwards using any method of your choice (remember that in the parentheses, you should get the above value if you solve for x):

If you multiply it out, you get (also your answer):
4. The zeros are at
x = 3, 2 and
-7.
Multiplicity of 3 is
1, for 2 it's
2, and for -7 it's
3.
Hope this helps!
H = 2sqrt2
D = 6sqrt3
In a 45,45,90 triangle the hypotenuse is (x)sqrt2 while the side lengths are equivalent being a single value x. Therefore, when given the hypotenuse and solving for the leg, divide 4 by sqrt2 to get 4sqrt2/2 which simplifies to 2sqrt2 when the denominator and numerator cancel.
In a 30,60,90 triangle the short leg x is across from the 30 degree angle meaning the angle across from the 60 degree angle is x times the sqrt of 3. Therefore the long leg is 6sqrt3
Answer:
C
Step-by-step explanation:
This problem is analogous to the extraction of 6 elements from a total of 10 elements. It's the same if they are marbles, chips, or in this case, people, as here we don't care about the order of the selection as we only are drawing a sample.
Thus, the problem implies solving the amount of possible combinations of 10 people if we take by 6. There is a formula for this and is:
10 C 6 = 10!(6!4!)
If we operate, knowing that for any number x, x!=x*(x-1)*(x-2)*...*1
10 C 6 = 10!(6!4!) = 10*9*8*7*6*...*1 / [(6*5*...*1) * (4*3*2*1)]
10 C 6 = 10*9*8*7*6! / [(6*5*...*1) * (4*3*2*1)]
We have a 6! multiplying and another dividing, so they get eliminated, and as 4*2=8 and 9=3*3
10 C 6 = 10*9*8*7*/ [(4*3*2*1)] = 10*3*3*8*7*/ [(8*3*1)]
We can eliminate the 8s and one of the 3s on the numerator with the one on the denominator:
10 C 6 = 10*3*7*/1 = 210/1= 210
So, option C