All categories

3 years ago

How you multiply 523x12

Mathematics

2 answers:

Mrac [35]3 years ago

3 0

You multiply 523x12 by writing it out. 523x12 is equal too 6,276

DedPeter [7]3 years ago

3 0

The answer to this is 6,276.

Hope it helps. :)

You might be interested in

I need your help guys please can you answer for me please

dybincka [34]

1. In parallelograms, opposite sides are congruent. Therefore EV=16
2. In parallelograms, adjacent angles add up to 180°, so measure of angle V = 100°
3. Opposite angles are equal so measure of angle L = 95°
4. Diagonals of parallelograms bisect each other. DE=10
5. Same rule as #4. LV=18

7 0

2 years ago

(a) Suppose you have 9 gold coins that look identical, but you also know one (and only one) of them is counterfeit. The counterf

alisha [4.7K]

Answer:

Step-by-step explanation:

You can split the coins into 3 groups, each of them has 3 coins. Weigh group 1 vs group 2, if one is lighter, that group has the fake coin. If both groups weigh the same, then group 3 has the fake coin.

Continue to split the group that has the fake coin into 3 groups, each group has 1 coin. Now apply the same procedure and we can identify the fake coin.

Total of scale usage is 2

b) if you have $3^n$ coins then you can apply the same approach and find the fake coin with just n steps. By splitting up to 3 groups each step, after each step you should be able to narrow down your suspected coin by 3 times.

Step 1: you narrow down to group of $\frac{3^n}{3} = 3^{n-1}$ coins

Step 2: you narrow down to group of $\frac{3^{n-1}}{3} = 3^{n-2}$ coins

Step 3: you narrow down to group of $\frac{3^{n-2}}{3} = 3^{n-3}$ coins

...

Step n: Step 1: you narrow down to group of $3^{n-n} = 3^0 = 1$ coin

8 0

3 years ago

A square has sides that are 80 ft long. Which equations can be used to calculate d the length of a diagonal of the square in fee

LenKa [72]

Notice the picture below

just use the pythagorean theorem to ge the diagonal

$\bf c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}$

7 0

3 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

3 years ago

4. A team of scientists is digging for fossils. The

STALIN [3.7K]

If it's not 6 cubic feet (because by the wording the answer is given).

If one dimension of the amount of soil they remove is 6, then you have to do 6*6*6= 216

If the total amount of soil removed is 6 cubic feet, then the dimensions of the cube would be $\sqrt[3]{6}$

6 0

2 years ago