Both problems give you a function in the second column and the x-values. To find out the values of a through f, you need to plug in those x-values into the function and simplify!
You need to know three exponent rules to simplify these expressions:
1)
The
negative exponent rule says that when a
base has a negative exponent, flip the base onto the other side of the
fraction to make it into a positive exponent. For example,

.
2)
Raising a fraction to a power is the same as separately raising the numerator and denominator to that power. For example,

.
3) The
zero exponent rule<span> says that any number
raised to zero is 1. For example,

.
</span>
Back to the Problem:
Problem 1
The x-values are in the left column. The title of the right column tells you that the function is

. The x-values are:
<span>
1) x = 0</span>Plug this into

to find letter a:

<span>
2) x = 2</span>Plug this into

to find letter b:

<span>
3) x = 4</span>Plug this into

to find letter c:

<span>
Problem 2
</span>The x-values are in the left column. The title of the right column tells you that the function is

. The x-values are:
<span>
1) x = 0</span>Plug this into

to find letter d:

<span>
2) x = 2
</span>Plug this into

to find letter e:

<span>
3) x = 4
</span>Plug this into

to find letter f:

<span>
-------
Answers: a = 1b = </span>

<span>
c = </span>
d = 1e =
f =
Answer:
I believe the numbers are 25 and 68.
Step-by-step explanation:
25 is the smaller number, 25 x 2 = 50, 50 + 18 = 68, and 68 + 25 = 93.
Given:
Radius of the circle = 10 in
Central angle of the sector = 45 degrees
To find:
The area of the sector.
Solution:
Area of a sector is

Where,
is the central angle in degrees.
Putting r=10 and
, we get



Therefore, the area of the sector is 12.5π sq. inches.
Answer:
√1370 or about 37.01
Step-by-step explanation:
use distance formula
√(x2-x1)^2+(y2-y1)^2
sqrt(8-7)^2+(41-4)^2
sqrt(1)^2+(37)^2
sqrt 1+1369
sqrt 1370
= about 37. 01
A, since the graph is not in a "U" shape aka parabola the degree cant be 2 so C is wrong and B and D are negative fuctions while the graph is showing one that is positive.