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fgiga [73]
3 years ago
11

Solve for g: 6g = 4 Need help on this problem??

Mathematics
2 answers:
seraphim [82]3 years ago
7 0
6g=4 now divide both sides by 6
g=4/6 simplify 
g=2/3
Harman [31]3 years ago
5 0
Remmber you can do anything to an equation as long as you do it to both sides

6g=4
divide both sides by 6
(6g)/6=4/6
(6/6)g=2/3
1g=2/3
g=2/3
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on Monday Mr. Borden ran 4.6 miles in the morning and 0.78 miles that afternoon. on Tuesday he ran 3.4 miles. how much did he ru
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X = total miles run

4.6 + 0.78 + 3.4 = x
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What is the least angle measure by which this figure can be rotated so that it maps onto itself?
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Whats the point of intersection between these two lines?<br><br>x+y=10<br>x+2y=14
Yakvenalex [24]
You have to make an equation that i forgot what was called but you just take
x+y=10 and multiply it by -2 so the y’s can cancel out

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Will all slices created by a slicing plane that passes through a sphere be the same size
Snowcat [4.5K]

Answer:

The suggestion proposed considering special cases, so that’s what we’ll do.

One special case we could consider is when x=0. If h=r, then we have a hemisphere. You may have learnt that the surface area of a sphere is 4πr2, so the answer in this case is 2πr2. Likewise, we can consider the most extreme cases: when h=2r, we get an area of 4πr2 (the entire sphere), and when h=0, we get zero area.

So if we do manage to get a general formula, we will be able to check it in these special cases.

Another possible idea when x=0 is to consider what happens when h is very small. In this case, the surface area looks a lot like a circle, but the radius of the circle does not seem particularly easy to work out, so maybe we’ll leave this one for a moment. (There are things we can do to approximate the radius using more advanced techniques, but they are currently beyond us.)

Another thing we could consider is the situation where h is very small. Then the part of the sphere between the planes looks very much like the frustum of a cone, and we know how to find the (surface) area of such a shape – see Cones.

If we recall our answer from that problem, we find the formulae for the surface area

π(R+r)s=π(R+r)hcosθ=π(R2−r2)sinθ.

Here, R is the radius of the base of the frustum, r is the radius of the top, h is the perpendicular height, s is the slant length and θ is the angle the slant makes with the vertical.

Step-by-step explanation:

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