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zubka84 [21]
3 years ago
7

A stackable CD rack holds 20 CDs and costs $3. Mark has a collection of 120 CDs. How much will it cost him to buy enough racks t

o hold all of his CDs?
Mathematics
1 answer:
elixir [45]3 years ago
5 0

Answer:

You would need $18 total to buy the racks needed.

Step-by-step explanation:

First, every 20 CDs needs a new rack, so you would divide 120 by 20.

120 / 20 = 6 racks

Second, you would multiply the price ($3) of a rack by how many racks needed.

3 * 6 = $18

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You are planning your cousin’s birthday party. He wants to invite at least 250 adults and children combined. Each adult will be
olga_2 [115]

The system of inequalities is x+y≥250 and 3x+2y≤800.

Step-by-step explanation:

Given,

People to invite = at least 250

At least 250 means that the number can increase from 250.

Cups to be provided = only 800

It means the number cannot exceed 800.

Number of adults = A

Number of children = C

According to given statement;

x+y≥250

3x+2y≤800

The system of inequalities is x+y≥250 and 3x+2y≤800.

Keywords: linear inequalities, addition

Learn more about linear inequalities at:

  • brainly.com/question/9738996
  • brainly.com/question/9729310

#LearnwithBrainly

5 0
3 years ago
the age of Jane is 80% of the age of Alice. If we add both ages the result is 45. Find the age of Jane and Alice
inysia [295]

Answer:

Age of Alice=25 years

Age of Jane=20 years

Step-by-step explanation:

We are given that the age of Jane is 80 % of the age Alice.

We have to find the age of Jane and Alice.

Let x be the age of Alice

According to question

Age of Jane=80% of Alice=80% of x=\frac{80}{100}\times x=\frac{4x}{5}

x+\frac{4x}{5}=45

\frac{5x+4x}{5}=45

\frac{9x}{5}=45

x=\frac{45\times 5}{9}=25

Age of Alice=25 years

Age of Jane=\frac{4}{5}\times 25=20 years

Age of Jane=20 years

5 0
4 years ago
At a local river there were 48 alligators laying on the river bank. If 5/6 of the alligators were sleeping, how many were NOT as
zhannawk [14.2K]

Answer:The answer is 40.

Step-by-step explanation:

8 0
3 years ago
A piece of wire 19 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
mr Goodwill [35]

Answer: 8.26 m

Step-by-step explanation:

$$Let s be the length of the wire used for the square. \\Let $t$ be the length of the wire used for the triangle. \\Let $A_{S}$ be the area of the square. \\Let ${A}_{T}}$ be the area of the triangle. \\One side of the square is $\frac{s}{4}$ \\Therefore,we know that,$$A_{S}=\left(\frac{s}{4}\right)^{2}=\frac{s^{2}}{16}$$

$$The formula for the area of an equilateral triangle is, $A=\frac{\sqrt{3}}{4} a^{2}$ where $a$ is the length of one side,And one side of our triangle is $\frac{t}{3}$So,We know that,$$A_{T}=\frac{\sqrt{3}}{4}\left(\frac{t}{3}\right)^{2}$$We have to find the value of "s" such that,$\mathrm{s}+\mathrm{t}=19$ hence, $\mathrm{t}=19-\mathrm{s}$And$$A_{S}+A_{T}=A_{S+T}$$

$$Therefore,$$\begin{aligned}&A_{T}=\frac{\sqrt{3}}{4}\left(\frac{(19-s)}{3}\right)^{2}=\frac{\sqrt{3}(19-s)^{2}}{36} \\&A_{T+S}=\frac{s^{2}}{16}+\frac{\sqrt{3}(19-s)^{2}}{36}\end{aligned}

$$Differentiating the above equation with respect to s we get,$$A^{\prime}{ }_{T+S}=\frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}$$Now we solve $A_{S+T}^{\prime}=0$$$\begin{aligned}&\Rightarrow \frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}=0 \\&\Rightarrow \frac{s}{8}=\frac{\sqrt{3}(19-s)}{18}\end{aligned}$$Cross multiply,$$\begin{aligned}&18 s=8 \sqrt{3}(19-s) \\&18 s=152 \sqrt{3}-8 \sqrt{3} s \\&(18+8 \sqrt{3}) s=152 \sqrt{3} \\&s=\frac{152 \sqrt{3}}{(18+8 \sqrt{3})} \approx 8.26\end{aligned}$$

$$The domain of $s$ is $[0,19]$.So the endpoints are 0 and 19$$\begin{aligned}&A_{T+S}(0)=\frac{0^{2}}{16}+\frac{\sqrt{3}(19-0)^{2}}{36} \approx 17.36 \\&A_{T+S}(8.26)=\frac{8.26^{2}}{16}+\frac{\sqrt{3}(19-8.26)^{2}}{36} \approx 9.81 \\&A_{T+S}(19)=\frac{19^{2}}{16}+\frac{\sqrt{3}(19-19)^{2}}{36}=22.56\end{aligned}$$

$$Therefore, for the minimum area, $8.26 \mathrm{~m}$ should be used for the square

8 0
2 years ago
It's going to be Lindsay's birthday soon, and her friends Mary, Jacqui, Tadeusz, Opal, and Tony have contributed equal amounts o
Soloha48 [4]
$30/5=6. Therefore, each person contributed 6 dollars
3 0
3 years ago
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