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dybincka [34]
3 years ago
14

Initially, there were only 86 weeds in the garden. The weeds grew at a rate of 29% each week. The following function represents

the weekly weed growth: f(x) = 86(1.29)x. Rewrite the function to show how quickly the weeds grow each day.
A.f(x) = 86(1.04)x; grows approximately at a rate of 0.4% daily
B.f(x) = 86(1.04)7x; grows approximately at a rate of 4% daily
C.f(x) = 86(1.29)7x; grows approximately at a rate of 20% daily
D.f(x) = 86(1.297)x; grows approximately at a rate of 2% daily
Mathematics
2 answers:
Orlov [11]3 years ago
7 0

Step-by-step answer:

The base of the exponential function is 1.29 for 7 days, as in

f(x) = 86*(1.29)^x

The new rate for days can be calculated by dividing x by 7 (where x remains the number of weeks), namely

f(x) = 86*1.29^(x/7)

Using the law of exponents, b^(x/a) = b^(x*(1/a)) = (b^(1/a))^x

we simplify by putting b=1.29, a=7 to get

f(x) = 86*(1.29^(1/7))^x

f(x) = 86*(1.037)^x      since 1.29^(1/7) evaluates to 1.037

Rounding 1.037 to 1.04 we get a (VERY) approximate function

f(x) = 86 * (1.04^x)

1.04 is very approximate because 1.04^7 is supposed to get back 1.29, but it is actually 1.316, while 1.037^7 gives 1.2896, much closer to 1.29.

mr_godi [17]3 years ago
7 0

Answer:

A

Step-by-step explanation:

i got it right on the test

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Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your
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The expression of integral as a limit of Riemann sums of given integral \int\limits^5_b {1} \, x/(2+x^{3}) dxis 4 \lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3} from i=1 to i=n.

Given an integral \int\limits^5_b {1} \, x/(2+x^{3}) dx.

We are required to express the integral as a limit of Riemann sums.

An integral basically assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinite data.

A Riemann sum is basically a certain kind of approximation of an integral by a finite sum.

Using Riemann sums, we have :

\int\limits^b_a {f(x)} \, dx=\lim_{n \to \infty}∑f(a+iΔx)Δx ,here Δx=(b-a)/n

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⇒Δx=(5-1)/n=4/n

f(a+iΔx)=f(1+4i/n)

f(1+4i/n)=[n^{2}(n+4i)]/2n^{3}+(n+4i)^{3}

\lim_{n \to \infty}∑f(a+iΔx)Δx=

\lim_{n \to \infty}∑n^{2}(n+4i)/2n^{3}+(n+4i)^{3}4/n

=4\lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3}

Hence the expression of integral as a limit of Riemann sums of given integral \int\limits^5_b {1} \, x/(2+x^{3}) dxis 4 \lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3} from i=1 to i=n.

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Answer: M∠2 just mean measure of angle 2,

Step-by-step explanation: So I worked it out and the awnser is B.

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Side 1 is 60 degrees and the left side is 20, so 20 + 60 + 3 (unkown    number) = 180

60 - 20 = 100,

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