Minor base: b=19 inches
Height: h=12.6 inches
Major base: B=29.2 inches
Area of the trapezoid: A
A=(b+B)h/2
Replacing the values:
A=(19 inches + 29.2 inches) (12.6 inches) / 2
A=(48.2 inches) (12.6 inches) / 2
A= (607.32 inches^2 ) /2
A= 303.66 inches^2
Answer: The area of the trapezoid is 303.66 square inches
The common ratio is 8
example:
-3 x 8 = -18
-18 x 8= -108
so on so on
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(
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x
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Move
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to the denominator using the negative exponent rule
b
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=
1
b
−
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.
⎛
⎝
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y
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⎞
⎠
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2
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Multiply
x
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by
x
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x
-
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by adding the exponents.
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(
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Move
y
−
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y
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2
to the numerator using the negative exponent rule
1
b
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n
=
b
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1
b
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b
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Multiply
y
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y
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by
y
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y
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by adding the exponents.
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⎛
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2
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2
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⎠
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Change the sign of the exponent by rewriting the base as its reciprocal.
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⎝
x
1
2
3
y
7
2
⎞
⎠
2
(
x
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2
Use the power rule
(
a
b
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n
=
a
n
b
n
(
a
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n
=
a
n
b
n
to distribute the exponent.
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(
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2
(
x
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Simplify the numerator.
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x
3
2
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2
x
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y
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2
Simplify the denominator.
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x
9
y
7
Yes the answer is correct
Answer:
Step-by-step explanation:
I'm sure you want your functions to appear as perfectly formed as possible so that others can help you. f(x) = 4(2)x should be written with the " ^ " sign to denote exponentation: f(x) = 4(2)^x
f(b) - f(a)
The formula for "average rate of change" is a.r.c. = --------------
b - a
change in function value
This is equivalent to ---------------------------------------
change in x value
For Section A: x changes from 1 to 2 and the function changes from 4(2)^1 to 4(2)^2: 8 to 16. Thus, "change in function value" is 8 for a 1-unit change in x from 1 to 2. Thus, in this Section, the a.r.c. is:
8
------ = 8 units (Section A)
1
Section B: x changes from 3 to 4, a net change of 1 unit: f(x) changes from
4(2)^3 to 4(2)^4, or 32 to 256, a net change of 224 units. Thus, the a.r.c. is
224 units
----------------- = 224 units (Section B)
1 unit
The a.r.c for Section B is 28 times greater than the a.r.c. for Section A.
This change in outcome is so great because the function f(x) is an exponential function; as x increases in unit steps, the function increases much faster (we say "exponentially").
