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vampirchik [111]
3 years ago
9

Super fine 40-gauge copper wire has a diameter of only 0.080mm and weighs only 44.5/gkm . Suppose a spool of 40-gauge wire weigh

s 205.g less after some wire is pulled off to wind a magnet. How could you calculate how much wire was used? Set the math up. But don't do any of it. Just leave your answer as a math expression. Also, be sure your answer includes all the correct unit symbols.
Mathematics
1 answer:
guajiro [1.7K]3 years ago
6 0

Answer:

Math expression=      L[m]=   W[g]   /   R[g/m]

answer=4.606Km

Step-by-step explanation:

It is known that the cable weighs 44kg / km therefore if the weight is divided by this amount we can find how much cable length was spent

L=W/R

L= leght of wire used(m)

W=weight used(g)=205g

R= relation between weight and leght(g/m)=44.5g/km=0.0445g/m

L=205g/0.0445g/m=  4606.74m=  4.606Km

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Solve the equation for y 

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4 0
3 years ago
The price P of a good and the quality Q of a good are linked.
Irina-Kira [14]

the equilibrium point, is when Demand = Supply, namely, when the amount of "Q"uantity demanded by customers is the same as the Quantity supplied by vendors.

That occurs when both of these equations are equal to each other.

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\bf \stackrel{\textit{Supply}}{-\cfrac{3}{4}Q+35}~~=~~\stackrel{\textit{Demand}}{\cfrac{2}{3}Q+1}\implies \stackrel{\textit{multiplying by 12}}{12\left( -\cfrac{3}{4}Q+35 \right)=12\left( \cfrac{2}{3}Q+1 \right)} \\\\\\ -9Q+420=8Q+12\implies 408=17Q\implies \cfrac{408}{17}=Q\implies \boxed{24=Q} \\\\\\ \stackrel{\textit{using the found Q in the Demand equation}}{P=\cfrac{2}{3}(24)+1}\implies P=16+1\implies \boxed{P=17} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{Equilibrium}{(24,17)}~\hfill



3 0
3 years ago
David keeps a range of stationery in his study. one could find pencils, erasers and pens in his study.5/7 of his stationery cons
tiny-mole [99]

The total units of stationery he has would be 55 approximately.

<h3>What is a system of equations?</h3>

A system of equations is two or more equations that can be solved to get a unique solution. the power of the equation must be in one degree.

David keeps a range of stationery in his study.

Let the pencil is represented by x.

Let the erasers be represented by y.

Let the pens be represented by z.

Let n be the stationery.

5/7 of his stationery consists of pencils.

x = 5/7 n

1/6 of the remaining stationery consists of erasers.

y = 1/6 n

There are 20 pens in his study.

z = 20

The total stationery

X + y + z

5/7 n + 1/6 n + 20 = n

20 = n -23/ 36 n

20 = 13 / 36 n

n = 55.38

Therefore, the total units of stationery he has would be 55 approximately.

Learn more about equations here;

brainly.com/question/10413253

#SPJ1

6 0
2 years ago
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