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aniked [119]
3 years ago
13

The value, V, of a certain investment doubles in every month, m. Which function can be used to find the value of the investment

at the end of m months if there is an initial value of $0.25?
V=0.25(m)2
cap v is equal to 0 point 2 5 times m squared

V=0.25(2)m
cap v is equal to 0 point 2 5 times 2 to the m th power

V=025(x)2m
cap v is equal to 025 times x raised to the 2 m power

V=2(0.25)m
Mathematics
1 answer:
Vladimir [108]3 years ago
4 0

bAnswer:

Step-by-step explanation:

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\displaystyle \int_0^3 f(x) \, dx = \frac12 \times 3 \times 5 = \boxed{\frac{15}2}

b) The integral is zero since the areas under the curve over [3, 4] and [4, 5] are equal but opposite in sign. In other words, on the interval [3, 5], f(x) is symmetric and odd about x = 4, so

\displaystyle \int_3^5 f(x) \, dx = \int_3^4 f(x) \, dx + \int_4^5 f(x) \, dx = \int_3^4 f(x) \, dx - \int_3^4 f(x) \, dx = \boxed{0}

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\displaystyle \int_5^9 f(x) \, dx = -4\times5 = -20

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