Is the answer <br> RATIO ? <br><br> *just checking*

zepelin [54]

Let C = the Chinese books.
Let E = the English language books.

C = 4/5 E

E - 300 = C - 180 Notice how you get this equation. You take 300 away from the English books and 180 from the Chinese and according to the question, you get an equal number of both. Put c = 4/5E into the second equation.

E - 300 = 4/5 E - 180 Add 300 to both sides.
E = 4/5 E - 180 + 300
E = 4/5 E + 120 Now subtract 4/5 E from both sidies.
E - 4/5 E = 120
1/5 E = 120 Now multiply by 5
E = 120 * 5
E = 600

So the number of books are
600 written in English
4/5 of 600 = 2400 /5 = 480 Chinese books.

English 600
Chinese 480

6 0

3 years ago

A runner increased her distance from 9 miles to 12.5 miles a week. Find the percent increase in mileage.

kondor19780726 [428]

Answer:

<u>39 percent</u> is increase in mileage.

Thus, the correct option is B. 39%.

Step-by-step explanation:

Given:

A runner increased her distance from 9 miles to 12.5 miles a week.

Now, to find percent increase in mileage.

Previous mileage = 9 miles.

Present mileage = 12.5 miles.

<em>Mileage increase = 12.5 - 9 = 3.5 miles.</em>

Now, to get the percent increase in mileage:

$\frac{Mileage\ increased}{previous\ mileage} \times 100$

$=\frac{3.5}{9} \times 100$

$=\frac{350}{9}$

$=38.88\%.$

<u><em>Approximately the percent increase = 39%.</em></u>

Therefore, 39% is increase in mileage.

Thus, the correct option is B. 39%.

4 0

3 years ago

A) Use the limit definition of derivatives to find f’(x)

Ann [662]

$\text{Given that,}\\\\f(x) = \dfrac{ 1}{3x-2}\\\\\text{First principle of derivatives,}\\\\f'(x) = \lim \limits_{h \to 0} \dfrac{f(x+h) - f(x) }{ h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3(x+h) - 2} - \tfrac 1{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3x+3h -2} - \tfrac{1}{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{3x-2-3x-3h+2}{(3x+3h-2)(3x-2)}}{h}\\\\\\$

$~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{-3h}{(3x+3h-2)(3x-2)}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{-3h}{h(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \lim \limits_{h \to 0} \dfrac{1}{(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \cdot \dfrac{1}{(3x+0-2)(3x-2)}\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)(3x-2)}\\\\\\~~~~~~~=-\dfrac{3}{(3x-2)^2}$

$\text{Given that,}~\\\\f(x) = \dfrac{1}{3x-2}\\\\\textbf{Power rule:}\\\\\dfrac{d}{dx}(x^n) = nx^{n-1}\\\\\textbf{Chain rule:}\\\\\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}\\\\\text{Now,}\\\\f'(x) = \dfrac{d}{dx} f(x)\\\\\\~~~~~~~~=\dfrac{d}{dx} \left( \dfrac 1{3x-2} \right)\\\\\\~~~~~~~~=\dfrac{d}{dx} (3x-2)^{-1}\\\\\\~~~~~~~~=-(3x-2)^{-1-1} \cdot \dfrac{d}{dx}(3x-2)\\\\\\~~~~~~~~=-(3x-2)^{-2} \cdot 3\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)^2}$