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Delvig [45]
3 years ago
5

Pls help me figure this out !!

Mathematics
1 answer:
scZoUnD [109]3 years ago
4 0

Answer:

I don't really see two or three points that are coplanar, so My guess would have to be either point G, E, or F.

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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
What is 5% increase of $40,000?
LuckyWell [14K]

Answer:

New value =

40,000 + Percentage increase =

40,000 + (5% × 40,000) =

40,000 + 5% × 40,000 =

(1 + 5%) × 40,000 =

(100% + 5%) × 40,000 =

105% × 40,000 =

105 ÷ 100 × 40,000 =

105 × 40,000 ÷ 100 =

4,200,000 ÷ 100 =

so the answer would be 42,000

3 0
3 years ago
Read 2 more answers
How long it takes a 2800 investment to earn $500 interest if it's invested at 7% compounded quarterly. Round to the nearest tent
OleMash [197]
P=present value of investment = 2800
i=interest per quarter=0.07/4=0.0175
n=number of quarters money invested (to be calculated)
Interest = P[(1+i)^n-1]=500
2800((1+0.0175)^n-1)=500
(1+0.0175)^n=500/2800+1
take log on both sides
n(log(1.0175)=log(1+500/2800)
n=log(1+500/2800)/log(1.0175)
=9.471 quarters
=37.88 months
=37.9 months [ nearest tenth of a month ]

4 0
3 years ago
Please help me online school isnt fit for me lol
worty [1.4K]

Answer:

The answer is A!!

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Can someone pleas3 help me with this, were doing algebraic expression I believe.
maxonik [38]

Answer:56w^{7} y^{5}\\

if it doesn't work im sorry

5 0
2 years ago
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