Ratio of modal boat to actual length of boat is 1:30.

Modal boat is 9 inches, therefore, the actual length is

inches, equal to 270 inches, or 22 feet, 6 inches.]]>I am looking for worksheets for this holiday season.

I am a tutor and with bunch of K5 kids, I want to make learning fun.

I have already got math worksheets for themes like Thanksgiving printable (https://logicroots.com/math-worksheets/thanksgiving-theme/) and Halloween printable (https://logicroots.com/math-worksheets/halloween-theme/).

I need for english as well. Any leads?

Welcome to the forum. Have you heard of the binomial distribution?

]]>Thanks a lot for all the help .]]>

When I've got a proof like this I find it helpful to consider an example first.

Let's say a = 48 = 2 x 2 x 2 x 2 x 3

and b = 56 = 2 x 2 x 2 x 7

The hcf = 2 x 2 x 2

and lcm = 2 x 2 x 2 x 2 x 3 x 7

so hcf x lcm = (2 x 2 x 2) x (2 x 2 x 2 x 2 x 3 x 7) = (2 x 2 x 2 x 2 x 3) x (2 x 2 x 2 x 7) = 48 x 56

The common factors occur in both the hcf and lcm and the not-common factors occur in the lcm. So the re-arrangement allows us to pick out one set of common factors together with one set of not-common factors for the first number and what is left is the factors for the other. Here's an attempt to make that rigorous:

Suppose a = hcf x N where N is the not-common factors, and similarly b = hcf x M

Then the lcm = all the common factors once and the not-common factors from both = hcf x N x M

So hcf x lcm = (hcf) x ( hcf x N x M) = (hcf x N) x ( hcf x M) = a x b

Hope that helps,

Bob

]]>Welcome to the forum.

Your reasoning is sound up until the last paragraph:

Since we are dealing with a limit process, we're interested when $m$ is a very large number, that is, when $m \gg 1$. In this particular, the second $+1$ in the first term above renders no effect. Therefore we can discard him.

multiply the first term top and bottom by y and the third by xy:

Bob

]]>Thanks ....I wonder how you could write the symbols of maths? is it the same sythax as lateX?

Yes.

]]>2

x= 5y

From that we can see that the minimum integer values ofxandyare5and2respectively (2 *5= 5 *2).

The minimum sum ofxandytherefore is5+2= 7.

Ugh... Why have I never thought about this?

]]>This result is true for any prism and cone which have the same cross-section / base. Curiously it's easier to prove in the general case than for a specific cross-section so I'll do that.

The first picture shows a prism which has a completely irregular, wiggly perimeter. The second shows the same cross-section; this time forming the base of a pyramid.

Firstly: a general principle for calculating a volume. ** If you know the area of a shape and then extend it in a third dimension at right angles to the base then the volume of this solid will be area of shape times the height of the extension.** Imagine the area divided into square centimetres. Each square makes a cuboid when extended so the result is true for these. If the shape is irregular, make as many square centimetres as you can, and then fit square millimetres for the rest. The result will be true for these square millimetres. Continue to sub-divide any remaining area into smaller and smaller squares and the result continues to be true. Hence it is true whatever the start area.

Now for the prism calculation:

Imagine the prism is divided into many small slices, each having the same cross-sectional area, A and 'height' delta x, a little bit in the x direction

Now for the pyramid.

The defining property for a pyramid is that the cross-section stays the same shape as the base but with a steadily diminishing size. If the base has area A, then a cross-section at height x (measured from the vertex) will have an area that is proportional to the area of the base. Say the base is at x = h and the vertex is at x = 0. The cross-section at height x is a shape that is similar to the base. The lengths are in the ratio h:x, so the areas will be in the ratio h^2 : x^2.

Hence

So the volume of the pyramid is one third the volume of the prism. The result in post 1 is just a special case where the cross-section is a circle.

Bob

]]>Bob

]]>If the four corners of a square touch the circumference of a circle then the diagonal of the square must be the diameter of the circle. So use Pythag. to calculate the diagonal and half it to get the radius.

If it's the other way round and the circle touches the sides of the square, then the length of the side is the diameter of the circle.

Bob

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