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Alexxandr [17]
3 years ago
5

Two math students were asked to write an exponential growth equation that had a starting value of 300 and a growth rate of 2%. P

ierre thinks the answer is y= 300 (1.02)x and Scott thinks that the answer is y= 300 (1.2)x Are either of them right and why?
Mathematics
1 answer:
Gemiola [76]3 years ago
5 0

Answer:

Pierre is right

Step-by-step explanation:

The correct formula for Exponential growth rate is given as:

y = a( 1 + r) ^t

Where

y = Amount after time t

a = Initial amount

r = Growth rate

t = time

From the question

a = 300

r = 2% = 0.02

Hence, our exponential growth rate =

y = 300( 1 + 0.02)^t

y = 300( 1.02)^t

Therefore, Pierre is right

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A television Mark 1600 was sold for 1200 what percent discount was given on that television​
adoni [48]

Answer:

25%

Step-by-step explanation:

Discount = 1600 - 1200 = 400

Percentage discount

=  \frac{400}{1600}  \times 100 \\  \\  =  \frac{400}{16}  \\  \\  = 25\%

5 0
3 years ago
Hey guys does anyone know how to lesson 26 Exit ticket 5.4 eureka math engage ny? The last problem: Larry spends half of his wor
Vesnalui [34]

Answer:

\frac{1}{12}X

Step-by-step explanation:

Let X is the number of hours Larry works a day

Given:  Larry spends half of his workday teaching piano lessons.

=> Number of hours he teaches piano a day is: \frac{1}{2}X

As we know that, he has 6 students, so the fraction of his workday is spent with each student is:

\frac{1}{2}X : 6

= \frac{1}{12}X

Hope it will find you well during the time of isolation.

7 0
3 years ago
Read 2 more answers
Write an algebraic expression for the word expression: <br> 15 divided by the sum of "d" and 3
Yuki888 [10]

the answer is

15/(3+d)

8 0
4 years ago
Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n
aliya0001 [1]

Answer:

f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...

We first compute the n-th derivative of f(x)=\ln(1+2x), note that

f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\

Now, if we compute the n-th derivative at 0 we get

f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!

and so the Maclaurin series for f(x)=ln(1+2x) is given by

f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2  \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}

3 0
3 years ago
Rectangle A has length 12 and width 8. Rectangle B has length 15 and width 10. Rectangle C has length 30 and width 15.
Anna007 [38]

Answer:

  yes; 1.25

Step-by-step explanation:

The length to width ratios of the rectangles are ...

  A: 12/8 = 1.5

  B: 15/10 = 1.5

  C: 30/15 = 2.0

__

Rectangles A and B have the same aspect ratio, so are similar. Rectangle B is a scaled copy of A with a scale factor of 10/8 = 1.25.

7 0
3 years ago
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