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Ronch [10]
3 years ago
7

I want help ,i want an answer

Mathematics
1 answer:
Vaselesa [24]3 years ago
8 0

Answer:

13) 196 - 49*pi

11) 9*pi

17) pi*(2.3/2)^2

19) 28/(4/9 + pi*(3/4)^2) where you only take the whole number and no decimals.  For instance if it was 3.99, the answer would be 3

Step-by-step explanation:

13)  So it looks like it is a shaded square with a blank circle placed on top and we want the area peeking out around the circle.  The trick here is to find the area of the square and circle, then see how much the circle is covering up by subtractign that fromt he square area.

square area: 14*14 = 196

for the circle you need the radius.  Well, you see int he middle of the circle it touches both ends of the square, this means it is also 14 across.  that's the diameter.  diamter is twice the radius so the radius is  14/2 = 7.  so the area is pi*7^2 = 49*pi

11)  You can imagine this as being similar to 13,  Iagine half a circle that then has a smaller semicircle laying on top of it so only that ribbon like part in the picture shows.  Again, we want to find the area of the big one and subtract the area of the small one.

for semicircles, since it is half a circle we just cut it in half.  Also it shows the radius of the inner crcle is 4 feet and the larger circle is one foot more, so 5 feet.  so area of the lage one is pi * 5^2 and area of the small one is pi * 4^2.  then subtracting the small from the large gets us 25*pi - 16*pi = (25-16)pi = 9*pi

17)  This has a lot of complicated words, but it just wants the area of a circle.  specifically the circle mentioned.  It gives us the diameter and we want the radius, the radius is 2.3/2.  so the area is pi * (2.3/2)^2

19)  This question is asking "what is the area f this figure, and when you know that we have this much paint, measured in an area.  How many of these pieces could we paint with this much paint."  if that is confusing here then.

Imagine you had a square of length 5 inches.  this means it has an area of 25 square inches.  if I then said I has 100 square inches of paint, how many squares could I paint?  basically eah time you paint a square it uses up 25 of that 100.  so after 4 it would have used up every drop exactly.

Now we do the same here.  find the area of the shape , and see how mnay times it could be painted with the paint we have.

the shape is made up of a rectangle and two semicircles.  Let's find the area of each part.

the rectangle is 1 1/3 long and 1/3 wide.  If you do not understand how I found those let me know.  I am going to write 1 1/3 as 4/3.  Now, area of a rectangle is just length times width.  so 4/3 * 1/3 = 4/9.

Now the semicircles.  they have the same measurements, so we just need to find one then multiply it by two.  It says the whole shape is 2 5/6 long, but the rectangle is 1 1/3 long, so that means the two semicircles mke up the rest.  so what plus that same thing plus 4/3 will equal 2 5/6 (I am going to write as 17/6)?  This is just algebra.  i am going to call the radii of the two semicircles x.

x + x + 4/3 = 17/6

2x = 17/6 - 4/3

2x = 17/6 - 8/6

2x = 9/6

x = 9/12

x = 3/4

So the radius of the semicircles is 3/4 feet.  now, I know I said we would find the area of one and double, but two semicircles makes one circle, so we can just find the area of ONE circle with that radius.  Kind of like we are pushing the two halves together.  

pi*(3/4)^2

Now, to find the area of the whole thing we add the area of the two semicircles we just found (or one regular circle) and the rectangle.  so 4/9 + pi*(3/4)^2

NOW we see how much 28 square feet can cover.  Just like in the example I gave you divide the amoutn you have by the area of the shape, so do that.  

28/(4/9 + pi*(3/4)^2)

You are probably going to get a number with a decimal.  Just take the whole number, the decimal part represents how much of one more shape you can paint.  So since it is not another whole other shape it doesn't matter for this answer.  

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Hope this helps have a nice nite
7 0
3 years ago
WHY IS THIS GETTING DELETED?? I NEED HELP
aliina [53]

If this exact question is repeatedly deleted, it's probably because of the ambiguity of the given equation. I see two likely interpretations, for instance:

\dfrac{(5\times5)^k}{5^{-8}} = 5^3

or

\dfrac{5\times 5^k}{5^{-8}} = 5^3

If the first one is what you intended, then

\dfrac{(5\times5)^k}{5^{-8}} = \dfrac{(5^2)^k}{5^{-8}} = \dfrac{5^{2k}}{5^{-8}} = 5^{2k-(-8)} = 5^{2k+8} = 5^3

and it follows that

2<em>k</em> + 8 = 3   ==>   2<em>k</em> = -5   ==>   <em>k</em> = -5/2

If you meant the second one, then

\dfrac{5\times 5^k}{5^{-8}} = \dfrac{5^1\times5^k}{5^{-8}} = \dfrac{5^{k+1}}{5^{-8}} = 5^{k+1-(-8)} = 5^{k+9} = 5^3

which would give

<em>k</em> + 9 = 3   ==>   <em>k</em> = -6

And for all I know, you might have meant some other alternative... When you can, you should include a picture of your problem.

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Step-by-step explanation:

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we have given the height radius of cylindrical can.

h=10cm r=10 cm .

and she have 6cm broth height in the tin that means she had used 4 cm of height of broth .

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and we need to find the difference between the volumes .

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volume 2=\pi *(10)^2*6=1884cm^3

volume used =volume 1-volume2=3140-1884=1256cm^3 is used

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