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3 years ago

Help Me Please. Im very stuck

Mathematics

1 answer:

Zielflug [23.3K]3 years ago

6 0

Answer:

28 sq units

Step-by-step explanation:

i wrote it down on paper so yea

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The new iPhone 3.14 is being sold at the UT Math Department. The original price is $1200 (day 0), but they are going to decrease

Marat540 [252]

$\bf \qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&50.14\\ P=\textit{initial amount}\dotfill &1200\\ r=rate\to r\%\to \frac{r}{100}\dotfill\\ t=\textit{elapsed time}\dotfill &16\\ \end{cases}$

$\bf 50.14=1200(1-r)^{16}\implies \cfrac{50.14}{1200}=(1-r)^{16}\implies \sqrt[16]{\cfrac{50.14}{1200}}=1-r \\\\\\ r=1-\sqrt[16]{\cfrac{50.14}{1200}}\implies r\approx 0.18\implies \stackrel{\textit{converting to percent}}{r\approx 0.18\cdot 100}\implies r\approx \stackrel{\%}{18}\quad \leftarrow x$

3 0

3 years ago

The average price of a gallon of gas increased by 12% in 2010 and by 10% in 2011. By what percentage did the average price of a

Leto [7]

Answer:

16.67% increase

Step-by-step explanation:

finding the percentage price increase

we need to find the difference in percentage price between the 2 years

12%-10% = 2%

2%/12% x 100% = 100%/6 = 16.67%

20% increase between January 1st 2010 and January 1st 2011

7 0

2 years ago

In a class of 29 students, 9 play an instrument and 23 play a sport. There are5

GenaCL600 [577]

Answer:

5/23

Step-by-step explanation:

23 play a sport and 5 play both an instrument and a sport so 5 out of 23 play an instrument given that they play a sport

3 0

2 years ago

1. (5pts) Find the derivatives of the function using the definition of derivative.

andreyandreev [35.5K]

2.8.1

$f(x) = \dfrac4{\sqrt{3-x}}$

By definition of the derivative,

$f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

We have

$f(x+h) = \dfrac4{\sqrt{3-(x+h)}}$

and

$f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}$

Combine these fractions into one with a common denominator:

$f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}$

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

$f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}$

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

$\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}$