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alexgriva [62]
3 years ago
14

I don’t know how to answer this please help

Mathematics
2 answers:
laila [671]3 years ago
8 0

Answer:

for the second square, 52% of it is shaded

hope this helps

LUCKY_DIMON [66]3 years ago
3 0

Answer:

I think 38%

Step-by-step explanation:

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PLEASEE HELP MEE ANSWER THIS​
frosja888 [35]
X=3
JM=9
the realationship is that it’s “splitting them in half”
3 0
3 years ago
Given A = 1 /2 (b1 + b2)h, solve for h.
Tatiana [17]

Answer: b: h=2a/b1+b2

Step-by-step explanation:

6 0
3 years ago
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
4.85 + c > - 2.55=what
VikaD [51]

Answer:

Step-by-step explanation:

Here you go mate

Step 1

4.85+c >-2.55  Equation/Question

Step 2

4.85 + c > - 2.55  Simplify

4.85 + c > - 2.55

Step 3

4.85 + c > - 2.55  add -4.85 to both sides

answer

c>-7.4

Hope this helps

4 0
3 years ago
A​ student's course grade is based on one midterm that counts as 5​% of his final​ grade, one class project that counts as 15​%
Darina [25.2K]

Answer:

  • 76.5
  • C

Step-by-step explanation:

Apply the weights to the scores and add them up:

  5% × midterm + 15% × project + 35% × homework + 45% × final

  = 0.05(75) +0.15(93) +0.35(78) +0.45(70)

  = 76.5

The student has a solid letter grade of C.

8 0
4 years ago
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