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4 years ago

What is the difference of (4-5y)-2(3.5y-8)?

Mathematics

1 answer:

Dafna1 [17]4 years ago

7 0

<span>20 - 12y Simplify parentheses: 4 - 5y - 7y + 16 Combine similar terms: 4 + 16 - 5y - 7y Simplify: 20 - 12y</span>

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Which equation in point slope form passes through the point (-1, 6) and has a slope of 2/3

marysya [2.9K]

Answer:

y - 6 = 2/3( x + 1)

Step-by-step explanation:

Equation for point slope form

y - y1 = m( x - x1)

( x1 , y1) - point on the line

m - slope of the line

Given

Slope: 2/3

Point ( -1 , 6)

x1 = -1

y1 = 6

m = 2/3

Substitute the value into the equation

y - 6 = 2/3(x - (-1)

y - 6 = 2/3( x + 1)

The equation of the line in point slope form is

y - 6 = 2/3( x + 1)

5 0

4 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

Create a matrix that is equal to F+G. The first matrix below is named F and the second matrix below is named G. Name the new mat

likoan [24]

F+G:

$F+G=\begin{bmatrix}{-1.8} & {-8.6} & {} \\ {2.85} & {-1.4} & {} \\ {-1.8} & {5.1} & {}\end{bmatrix}+\begin{bmatrix}{1.32} & {-1.9} & {} \\ {2.25} & {0.0} & {} \\ {-6.2} & {1.4} & {}\end{bmatrix}$

Then, add the elements that occupy the same position: