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ICE Princess25 [194]

3 years ago

14

Mr. Nelson has a class of 16 students. He can spend $22 on each student to buy math supplies for the year. He first buys all of

his students calculators, which costs a total of $117.76. After buying the calculators, how much does he have left to spend on each student?

2 answers:

Nuetrik [128]3 years ago

6 0

Answer:

$14.64

Step-by-step explanation:

First, find how much he spent on the calculators

Divide the amount spent on calculators by the number of students

calculator cost/students

117.76/16=7.36

So, he spent 7.36 on each student so far.

To find how much he still has left to spend, subtract the amount he has spent from the total he can spend

total-$ spent on calculators

22-7.36=14.64

He can still spend $14.64 on each student

Hope this helps! :)

Yuri [45]3 years ago

3 0

If he can spend $22 per student, that means he can spend a total of $352 when you multiply 22 times 16. Because he spent $117.76, that means he has a remainder of $234.24 to spend on his students. Because he has 16 students and needs to divide the money among them equally, 234.24/16 means there is $14.64 to spend on each student.

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3 years ago

The total number of people at a football game was 5600. Field-side tickets were 40 dollars and end-zone tickets were 20 dollars.

rewona [7]

Answer:

1100 field-side tickets and 4500 end-zone tickets.

Step-by-step explanation:

Let x represent number of field side tickets and y represent number of end-zone tickets.

We have been given that the total number of people at a football game was 5600. We can represent this information in an equation as:

$x+y=5600...(1)$

$y=5600-x...(1)$

We are also told that Field-side tickets were 40 dollars and end-zone tickets were 20 dollars.

Cost of x field side tickets would be $40x$ and cost of y end-zone tickets would be $20y$ .

The total amount of money received for the tickets was $134000. We can represent this information in an equation as:

$40x+20y=134000...(2)$

Upon substituting equation (1) in equation (2), we will get:

$40x+20(5600-x)=134000$

$40x+112000-20x=134000$

$20x+112000=134000$

$20x+112000-112000=134000-112000$

$20x=22000$

$\frac{20x}{20}=\frac{22000}{20}$

$x=1100$

Therefore, 1100 field side tickets were sold.

Upon substituting $x=1100$ in equation (1), we will get:

$y=5600-1100$

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3 years ago

Kathy has $2 less than 3 times the amount of money jason has. Together, they have a total of $34. How much money does Kathy has?

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3 years ago

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru

Viktor [21]

Answer:

1) $S = 2\cdot w\cdot l - 8\cdot x^{2}$ , 2) The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$ , 3) $S = 176\,in^{2}$ , 4) $x \approx 4.528\,in$ , 5) $S = 164.830\,in^{2}$

Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

3) The surface area when a 1'' x 1'' square is cut out is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

$x \approx 4.528\,in$

5) The equation of the box volume is:

$V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x$

$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

$V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}$

$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

The first derivative of the function is:

$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

The critical points are determined by equalizing the derivative to zero:

$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

$x_{1} \approx 4.952\,in$

$x_{2}\approx 1.548\,in$

The second derivative is found afterwards:

$V'' = 24\cdot x - 78\,in$

After evaluating each critical point, it follows that $x_{1}$ is an absolute minimum and $x_{2}$ is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

$x \approx 1.548\,in$

The surface area of the box is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}$

$S = 164.830\,in^{2}$

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