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ladessa [460]
1 year ago
13

nd the surface area of a cylinderose radius is 1.2 mm and whoseheight is 2 mm.Round to the nearest tenth.SA = [?] mm²

Mathematics
1 answer:
AleksAgata [21]1 year ago
3 0

Step 1

Given;

\begin{gathered} A\text{ cylinder with radius 1.2mm} \\ Height=2mm \end{gathered}

Required; To find the area

Step 2

The area of a cylinder is given as;

\begin{gathered} A=2\pi rh+2\pi r^2 \\ A=(2\times\pi\times1.2\times2)+(2\times\pi\times1.2^2) \\ A=4.8\pi+2\times\:1.2^2\pi \\ A=24.12743mm^2 \\ A\approx24.1mm^2 \end{gathered}

Answer;

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Deffense [45]

Answer: the answer is 8

Step-by-step explanation:

-6 + 8 = 2 (y)

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3 years ago
How many committees of 4 boys and 3 girls<br> can be formed from a class of 6 boys and 7<br> girls?
VLD [36.1K]

Answer:

525

Step-by-step explanation:

This is a question involving combinatorics

The number of ways of choosing a subset k from a set of n elements is given by {n \choose k} which evaluates to \frac{n!}{k!(n-k)!}

n! is the product n × (n-1) × (n-2) x....x 3 x 2 x 1

For example,

4! = 4 x 3 x 2 x 1 = 24

3! = 3 x 2 x 1 = 6

Since we have to choose 4 boys from a class of 6 boys, the total number of ways this can be done is

{6 \choose 4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!}

Note that 6! = 6 x 5 x 4 x 3 x 2 x 1 and 4 x 3 x 2 x 1  is nothing but 4!

So the numerator can be re-written as 6 x 5 x (4!)

We can rewrite the expression \frac{6!}{4!2!} \text{ as } \frac{6.5.4!}{4!2!}

Cancelling 4! from both numerator and denominator gives us the result

as  (6 × 5)/2! = 20/2 = 15 different ways of choosing 4 boys from a class of 6 boys

For the girls, the number of ways of choosing 3 girls from a class of 7 girls is given by

{7 \choose 3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!}

This works out to (7 x 6 x 5 )/(3 x 2 x 1)  (using the same logic as for the boys computation)

= 210/6 = 35

So total number of committees of 4 boys and 3 girls that can be formed from a class of 6 boys and 7 girls = 15 x 35 = 525

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2 years ago
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Answer:

Step-by-step explanation:

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If these are the length and width of the smaller one, we can multiply each by 5/4 to get the length and width of the larger one.

length of larger rectangle = 60*5/4 = 75

width of larger rectangle = 40*5/4 = 50

These dimensions would give a perimeter of 250, since 2*(50 + 75) = 250.

It is worth noticing that 250 is equal to 200 * 5/4. In the future, you can simply multiply perimeter by the ratio instead of going through all these steps.

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3 years ago
Please answer the middle
alukav5142 [94]

Table-1 shows constant increase of X and Y values

Step-by-step explanation:

We are given two tables with X and Y values.

To find which table represent the constant increase of X and Y:

If there is constant increase of X and Y then, X and Y avlues belongs to equation of line.

Also, if three points lies on line then, slope of line made by three points must be same or equal.

Slope of line is given as s=\frac{Y2-Y1}{X2-X1}

For table 1:

The values of X and Y form points as A(3,4),B(5,6) and C(7.8)

For A(3,4) and B(5,6)

s=\frac{6-4}{5-3}

s=\frac{2}{2}

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For C(7.8) and B(5,6)

s=\frac{6-8}{5-7}

s=\frac{-2}{-2}

s=1

Therefore, Table-1 shows constant increase of X and Y values

For table 2:

The values of X and Y form points as A(1,3),B(2,6) and C(3,10)

For A(1,3) and B(2,6)

s=\frac{6-3}{2-1}

s=\frac{3}{1}

s=3

ForC(3,10) and B(5,6)

s=\frac{6-10}{5-3}

s=\frac{-4}{-2}

s=2

Therefore, Table-1 does not shows constant increase of X and Y values

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3 years ago
Indicate the equation of the given line in standard form.
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7 0
4 years ago
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