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hammer [34]
3 years ago
13

An elementary school teacher bought a bag of candy and brought it to school as a reward for the entire class, planning to give i

t out at the end of school day. The teacher counted the candy ahead of time and found that there would be 7 left over if they gave each of their 25 students the same number of pieces, so they ate 7 pieces before school started. However, only 21 students showed up to school that day. Since dividing the candy evenly would now result in 16 leftover pieces, the teacher left 16 pieces in the teachers lounge during lunch. 2 more students went home during recess, resulting in a final total of 8 pieces left over after the candy was divided evenly at the end of the day. What is the smallest number of candies the teacher could have started with
Mathematics
2 answers:
True [87]3 years ago
6 0

Answer:

divid than add

Step-by-step explanation:

dmitriy555 [2]3 years ago
4 0
You divide and then you add
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Derivative of<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%7B3x%7D%5E%7B2%7D%20-%202x%20-%201%20%7D%7B%20%7Bx%7D%5E%7B2
Anastaziya [24]

Answer:

\displaystyle  \frac{dy}{dx} =    \frac{2x + 2}{x^3}

Step-by-step explanation:

we would like to figure out the derivative of the following:

\displaystyle  \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} }

to do so, let,

\displaystyle y =  \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} }

By simplifying we acquire:

\displaystyle y =  3 -  \frac{2}{x}  -  \frac{1}{ {x}^{2} }

use law of exponent which yields:

\displaystyle y =  3 -  2 {x}^{ - 1}  -   { {x}^{  - 2} }

take derivative in both sides:

\displaystyle  \frac{dy}{dx} =  \frac{d}{dx}  (3 -  2 {x}^{ - 1}  -   { {x}^{  - 2} } )

use sum derivation rule which yields:

\rm\displaystyle  \frac{dy}{dx} =  \frac{d}{dx}  3 -   \frac{d}{dx} 2 {x}^{ - 1}  -     \frac{d}{dx} {x}^{  - 2}

By constant derivation we acquire:

\rm\displaystyle  \frac{dy}{dx} =  0 -   \frac{d}{dx} 2 {x}^{ - 1}  -     \frac{d}{dx} {x}^{  - 2}

use exponent rule of derivation which yields:

\rm\displaystyle  \frac{dy}{dx} =  0 -   ( - 2 {x}^{ - 1 -1} ) -     ( - 2 {x}^{  - 2 - 1} )

simplify exponent:

\rm\displaystyle  \frac{dy}{dx} =  0 -   ( - 2 {x}^{ -2} ) -     ( - 2 {x}^{  - 3} )

two negatives make positive so,

\displaystyle  \frac{dy}{dx} =   2 {x}^{ -2} +      2 {x}^{  - 3}

<h3>further simplification if needed:</h3>

by law of exponent we acquire:

\displaystyle  \frac{dy}{dx} =   \frac{2 }{x^2}+       \frac{2}{x^3}

simplify addition:

\displaystyle  \frac{dy}{dx} =    \frac{2x + 2}{x^3}

and we are done!

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