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Over [174]
3 years ago
14

One ton=2,000 pounds. In pounds, what is the weight of each cylinder?

Mathematics
1 answer:
jonny [76]3 years ago
7 0
I'm pretty sure its 1,000 pounds
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How do you solve simple linear equations
Vera_Pavlovna [14]
It’s easy just make a line and put numbers each line to solve the linear equation and than solve it simple
4 0
3 years ago
Read 2 more answers
Jenny wants to rent a truck for one day. She contacted two companies. Laguna’s Truck Rentals charges $20 plus $2 per mile. Salva
PolarNik [594]
20+2x=3x subtract 2x from both sides
20=x after 20 miles the total cost for both companies will be the same

PROOF
20+2(20)=3(20)
20+40=60
60=60
3 0
3 years ago
A farmer wants to build a rectangular pen and then divide it with two interior fences. The total area inside of the pen will be
irina [24]

Answer:

Dimensions of the pen:

x = 11 ft

y = 6 ft

Step-by-step explanation:

Let call "x"  and "y"  dimensions of the rectangular pen  and  x > y, so the interior fences will be equal to y.

The exterior length is 2*x + 2*y and its cost is (2*x + 2*y ) *20.40

The interior fences are 2*y  and its cost is 2*y* 17

Total cost C  = (2*x + 2*y ) *20.40 +  2*y* 17       (1)

Now area inside the pen is 66 ft²  and it is equal to:

A = x*y    ⇒  66 = x*y   ⇒  y = 66/x

Plugging that value in equation (1) will give C as a function of x

C(x) = [ 2*x + 2* (66/x) ]* 20.40  +  2* (66/x) * 17

C(x) = ( 2*x + 132/x ) 20.40  +  2244/x

C(x)  = 40,80*x + 2692.8/x  + 2244/x

C(x) = 40.80*x  + 4936,8/x

Taking derivatives on both sides of the equation we get

C´(x)  = 40.80 - 4936,8/x²

C´(x)  = 0    ⇒   40.80 - 4936,8/x² = 0   ⇒  40.80 *x² = 4936,8

x² =  4936,8 / 40.80    ⇒  x²  = 121  ⇒  x √121

x = 11 ft

And  y = 66/x   ⇒  y  = 66/11    ⇒  y = 6 ft

3 0
4 years ago
Two cars leave Denver at the same time and travel in opposite directions. One car travels 10 mi/h faster than the other car. The
mestny [16]
Recall your d = rt, distance =  rate * time

let's say we have two cars, A and B, B has a speed rate of "r", so A is the faster one and has a speed rate of " r + 10 "

they both are 300 miles apart after 3hrs, after 3hrs, car has been running for 3hrs, and car B has also been running for 3hrs, so their time is the same

we know their distances added up, is 300miles, so if car B covered say, "d" distance on those 3hrs, car A covered the slack, " 300 - d "

\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{car A}&300-d&r+10&3\\
\textit{car B}&d&r&3
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
300-d=(r+10)3\\
\boxed{d}=3r\\
----------\\
300-\boxed{3r}=3(r+10)
\end{cases}

solve for "r"
8 0
4 years ago
PLEASE HELP ANSWER GETS
Ksenya-84 [330]

Answer:

i don't know, it's either C or B

Step-by-step explanation:

8 0
4 years ago
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