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nikitadnepr [17]
3 years ago
14

2. Mr. Andrew wants to give each of his students an equal number of the crayons he bought. There are 32 students in Mr. Andrew’s

class. How many crayons will each student get?
Estimate first.





Solve the problem and explain your answer using pictures, diagrams, equations, and/or words.




3. How many more boxes of crayons does Mr. Andrew need if he wants each of his students to get 12 crayons? Explain your answer.
Mathematics
2 answers:
Natali [406]3 years ago
5 0

Answer:

It’s confusing

Step-by-step explanation:

Vladimir79 [104]3 years ago
3 0

Answer:

Step-by-step explanation:

2.

Because we don't know how many crayons he bought we will call it c.

c = total number of crayons purchased

T = total number of crayons each student gets

To determine how many crayons each student gets we divide c by 32, because that's the number of students in his class.

c / 32 = T

So let's say he purchased 120 crayons.

c = 120

120/32 = T

T = 3.75

Since he can't give a fraction of a crayon, in this case, each student will get 3 crayons.

3.

We will now pretend there are 24 crayons in 1 box.

So T still equals the total number of crayons each student gets and c still equals number of crayons purchased, but now we also have b which equals the number of crayons in 1 box.

c = total number of crayons purchased

T = total number of crayons each student gets

b = number of crayons in 1 box

x = number of boxes purchased

So b*x = c

We can still write the equation:

c / 32 = T

We know T = 12 So we plug 12 in for T

c / 32 = 12  Now multiply each side by 32, so 32 will cancel on the left.

32c/32 = 12 * 32

c = 12 * 32

c = 384

So if he needs to purchase 384 crayons we can use our other equation.

b*x = c    

b = 24

c = 384

So we plug those in.

24x = 384  Divide each side by 24 so 24 will cancel on the left.

24x/24 = 384/24

x = 384/24

x = 16

So he needs to buy 16 boxes of crayons in this instance. We can figure out how many boxes he bought in #2 by using the same equation.

b*x = c  

c = 120 in #2

b =24

24 * x = 120

24x/24 = 120/24

x = 120/24

x = 5

So for each student to have 12 crayons he needs to purchase 11 more boxes (16 - 5 = 11).

This will work for whatever numbers you were given.

Remember:

c = total number of crayons purchased

T = total number of crayons each student gets

b = number of crayons in 1 box

x = number of boxes purchased

b*x = c

c / 32 = T

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trapecia [35]

For this case we have the following fractions:

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We must rewrite the fractions, using the same denominator.

We have then:

We multiply the first fraction by 11 in the numerator and denominator:

\frac {11} {2} \frac {11} {11}

We multiply the second fraction by 2 in the numerator and denominator:

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Rewriting we have:

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For the second fraction:

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\displaystyle\\Answer:\ x=\sqrt{2}\ \ \ \ \ x=-\sqrt{2}\ \ \ \ \ x=\frac{ln5}{2}

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5x^2-x^2e^{2x}+2e^{2x}=10

Let:\ x^2=t  \ \ \ \ \ e^{2x}=v\\Hence,\\5t-tv+2v^2=10\\5t-tv+2v^2-10=10-10\\5t-tv-10+2v^2=0\\t(5-v)-2(5-v)=0\\(5-v)(t-2)=0\\5-v=0\\5-v+v=0+v\\5=v\\5=e^{2x}

<em>Prologarithmize both parts of the equation:</em>

ln5=lne^{2x}\\ln5=2x*lne\\ln5=2x*1\\ln5=2x\\

<u><em> Divide both parts of the equation by 2:</em></u>

\displaystyle\\x=\frac{ln5}{2}

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