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soldier1979 [14.2K]
3 years ago
6

Brainliest if correct

Mathematics
1 answer:
n200080 [17]3 years ago
6 0

Answer:

16 and 125

Step-by-step explanation:

-2*-2 = 4

4*-2 = -8

-8*-2 = 16

5*5 = 25

25*5 = 125

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Which fraction is NOT equivalent to 8/12<br> A. 2/3<br> B. 24/36<br> C. 4/10<br> D. 6/10
Hatshy [7]

Answer:

C. 4/10  ----> NOT equivalent to 8/12

D. 6/10  ---> NOT equivalent to 8/12


Step-by-step explanation:


A. 2/3  ----> equivalent to 8/12

B. 24/36  ----> equivalent to 8/12

C. 4/10  ----> NOT equivalent to 8/12

D. 6/10  ---> NOT equivalent to 8/12

5 0
3 years ago
Given the x intercept of (4,0) and y intercept of (0,-3) graph the points and find the slope of the line
Marina CMI [18]

Answer:

The slope is 3/4 and on a graph it would be 4 over to the right and 3 down.

4 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
Divide these polynomial a (27x 4 - 18x 3 + 9x 2) ÷ 3x
Blizzard [7]
The answer is 9x^3-6x^2+3x
5 0
3 years ago
Please answer correctly, will give brainliest
larisa [96]

Answer:

b

Step-by-step explanation:

7 0
3 years ago
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