The length of a beige is 500 miles if 1 inch represent 25 miles what is the length of the bridge ?

Gizmo can eat 2 bowls of kibbles in 3 minutes. Leo can eat one bowl of kibbles in 6 minutes. Together, how many bowls of kibbles

Westkost [7]

Gizmo eats two bowls in 3 minutes, which means in a minute he would eat a fraction of 2/3 of the bowl.
On the other hand, leo eats one bowl in 6 minutes hence in a minute he eats a fraction of 1/6 of a bowl.
When both combine, then in one minute they can eat a fraction of 2/3+1/6= 5/6.
Therefore, in ten minutes they will take 10× 5/6 = 8.333 equivalent to 8 bowls and a third of a bowl

4 0

2 years ago

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What ordered pair is the solution to -7x+3y=2

goldenfox [79]

Answer:

Step-by-step explanation:

3y=7x+2

$y=\frac{7x+2}{3}\\x=1,y=3\\$

one pair is (1,3)

go on adding 3 in x=1,you get corresponding value of y.

x=4,y=10,(4,10)

and so on .

7 0

3 years ago

A hollow metal sphere has 6 cm inner and 8cm outer radii. The surface charge densities on the exterior surface is +100 nC/m2 and

natulia [17]

Answer:

<h2>Outer Electric Field is 11250 N/C.</h2><h2>Inner Electric Field is -10000 N/C.</h2>

Step-by-step explanation:

First of all, we need to read carefully and analyse the problem. As you can see, is an electrical subject, and it's given surface charge densities and radius.

So, to calculate electric fields, we need to find the proper equation to do so: $E=k\frac{q}{r^{2} }$ ; as you can see, first we need to find the charges.

We can find all charges using the surface charge densities, because it has the next relation: $p=\frac{q}{A}$ ; which indicates that charge density is the amount of charge per area. But, there's a problem, we don't have areas, so we have to calculate them first with this relation: $S=4\pi r^{2}$ ; which gives us the surface of a sphere.

The inner surface: $Si=4\pi (0.06m)^{2} = 0.04 m^{2}$

The outer surface: $S=4\pi (0.08m)^{2}=0.08m^{2}$

Now we can calculate the charges,

Inner charge: $Qi=pA=(-100\frac{nC}{m^{2} } )(0.04m^{2} )=-4nC$

Outer charge: $Qo=pA=100\frac{nC}{m^{2} } )(0.08m^{2} )=8nC$

Then, we are able to calculate both fields:

Inner field: $Ei=k\frac{Qi}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{-4x10^{-9} }{0.06m^{2} }=-10000\frac{N}{C}$

Outer field: $Eo=k\frac{Qo}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{8x10^{-9} }{0.08m^{2} }=11250\frac{N}{C}$

The directions that field have is opposite each other, the inner one has an inside direction, and the outer electric field has an outside direction.

3 0

3 years ago

At 1:00 pm the water level in a pool is 13 inches. At 1:30pm the water level is 18 inches. At 2:30pm the water level is 28 inche

SashulF [63]

From the given question we come to know of certain number of facts and they are:
At 1:00 PM the water level of the pond was = 13 inches
At 1:30 PM the water level of the pond was = 18 inches
At 2:30 PM the water level of the pond was = 28 inches
From the above given facts we can easily find the amount of water changing every half an hour.
Amount of increase in water from 1:00PM to 1:30 PM = (18 - 13) inches
= 5 inches
Amount of increase in water level from 1:30PM to 2:30PM = (28 -18) inches
= 10 inches
From the above two deductions we can come to the conclusion the the constant rate of change in water level is 5 inches for every half an hour.

4 0

3 years ago

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Which information is necessary to solve this problem?

prohojiy [21]

That would be 400 / 28 = 14.29 gallons to nearest hundredth

3 0

3 years ago