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blsea [12.9K]
3 years ago
11

Four classmates were asked to decorate one fifth of the bulletin board in a classroom. They divided the section to be decorated

equally among 4 classmates. What fraction of the entire bulletin board did each classmate decorate
Mathematics
2 answers:
lara31 [8.8K]3 years ago
5 0
(1/5) x (1/4) = 1/20
each classmate decorated 1/20th of the entire board
Nat2105 [25]3 years ago
3 0

Answer:120

Step-by-step explanation: 1/5 x 1/4 = 120.

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Un granjero tiene 100 vacas que pesan 125 kg cada una. El costo diario de mantenimiento de una vaca asciende a $ 5.00. Las vacas
solniwko [45]

Answer:

25 days

Step-by-step explanation:

Assuming that the farmer waits for n days to get get the maximum profit.

Given that the total number of cows the farmer has, = 100

Currect weight of one cow = 125 kg.

Weight gain rate for one cow = 3 kg/day

So, weight gained by one cow in n days = 3n kg

Therefore, the weight of one cow after n days = 125 + 3n \;\;kg \cdots(i)

Cost for keeping one cow = $5/day.

Cost for keeping one cow for n days = $ 5n

So, Cost for keeping 100 cows for n days = \$ 5n \times 100 =\$500n \cdots(ii)

Current market price = 25 pesos / kg = 125 cents / kg [ as 1 peso = 5 cents]

The falling rate of market price = 1 cent /day.

Fall in price in n days = 1\times n = n cents.

So, market price after n days = 125-n cent/kg\cdots(iii)

By using equations (i) and (iii),

The selling price of one cow after n days = (125+3n) \times (125-n) cents.

So, the selling price of 100 cows after n days = (125+3n) \times (125-n)\times 100 cents.

As 1 $ = 100 cents. so

The selling price of 100 cows after n days =\$ 5(125+3n) (125-n)\cdots(iv).

Now, from equations (ii) and (iv)

Net profit, P = 5(125+3n) (125-n) - 500n\cdots(v)

To get the maximum profit, differentiate the profit function in equation (v) with respect to n and equate it to zero to get the value of n, i.e

\frac {dP}{dn}=0 \\\\\Rightarrow \frac {d}{dn}(5(125+3n) (125-n) - 500n)=0 \\\\\Rightarrow  5[ (125+3n)(-1)+(125-n)(3)]-500=0 \\\\\Rightarrow 5[ -125-3n+375-3n]-500 =0 \\\\\Rightarrow 5[ -125-3n+375-3n]=500\\\\\Rightarrow 250-6n=500/5=100 \\\\\Rightarrow 6n = 250-100=150 \\\\\Rightarrow n= 150/6 = 25.

Hence, the farmer has to wait for 25 days to get the maximum profit.

3 0
3 years ago
M&lt;ZHG = 11x - 1, m&lt;ZIHZ=24°<br>and m&lt;IHG = 12x + 13. Find m&lt;IHG.​
nasty-shy [4]

Answer:

m<IHG=133°

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

m<ZHG+m<IHZ= m<IHG -----> by angle addition postulate

substitute the given values  and solve for x

11x-1+24=12x+13

Combine like terms

11x+23=12x+13

Group terms that contain the same variable

12x-11x=23-13

x=10

Find the measure of angle IHG

substitute the value of x

m<IHG= 12(10)+13 =133°

8 0
4 years ago
Someone also help me with this one
Gre4nikov [31]

2m

I think it is because the down it is 8 m

so there is 6m it the top

so it should be

8-6=2

6 0
4 years ago
Item 5<br> Evaluate the expression.<br><br> |12(16−63)|
USPshnik [31]
564 is the answer here you go
6 0
4 years ago
At what rate was an investment made that obtains $125.37 on $597 over three years?
Serjik [45]
I use simple annual interest to solve the problem

<span>$125.37 -------------------------------- 3 years
X-------------------------------------------> 1 year
X=125.37/3=</span>$41.79/ year

<span>$597--------------------->100%
</span>$41.79---------------------X%

X=41.79*100/597=7%
 The rate was 7%
5 0
4 years ago
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