All categories

3 years ago

Whats the área and circumference if the diameter is 10mm?

Mathematics

2 answers:

elena-14-01-66 [18.8K]3 years ago

7 0

Answer:

78.54

Step-by-step explanation:

zmey [24]3 years ago

7 0

Answer:

Circumference of a circle in terms of radius:

Circumference = 2·π·r = 2·3.14·5 = 31.42 mm.(*)

In terms of diameter:

Circumference = π·d = 3.14·10 = 31.4 mm.(*)

In terms of area:

Circumference C = √4·π·A = √4·π·78.54 = 31.42 mm.(*)

Step-by-step explanation:

Note: For simplicity, the operations above were rounded to 2 decimal places and π was rounded to 3.14.

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It takes most oceans about 6 hours to go from high tide to low tide.Suppose that the water level decreases at a rate of 3 feet p

juin [17]

Answer:

From the level of high tide to low tide, it's 18 feet.

Step-by-step explanation:

To find the change in water level, we multiply the rate (3 ft / hr) by the time (6 hrs), obtaining 18 ft.

The water level changes by 18 ft.

8 0

2 years ago

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

2 years ago

Two puppies weigh 4/5 of a kilogram. if one puppy weighs 2/3 of a pound what is the weight if the other puppy?

jeyben [28]

Answer:

2/15

Step-by-step explanation:

4/5-2/3

12/15-10/15

2/15

8 0

2 years ago

What are the lengths of DE and EF. Answer with supporting work:

nataly862011 [7]

The length of DE is 39 and the length of EF is 36 because we can see from side AC and DF that triangle DEF is just triangle ABC dilated by 3.

5 0

3 years ago

Hello please help here!! thank you!!

Rudiy27

Answer:

x=110

Step-by-step explanation:

Since the sum of the interior angles of a trapazoid is 360, all you have to do is add the angles given and subtract that by 360, which gets you 110.

3 0

3 years ago