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3 years ago

Prove that the value of the expression: 165+164 is divisible by 17

Mathematics

2 answers:

galina1969 [7]3 years ago

5 0

Answer:

329 divided by 17 is 19.3529411765 so round it 19.3529 turns in to 19.353

so take that and mutiply it by 17 then you will get 329

Step-by-step explanation:

Zepler [3.9K]3 years ago

4 0

Answer:

$16^{2} (272)$ is divisible by 17

Proof:

In order to solve this question we will have to factor $16^{5} + 16^{4}$ . We will have to do that as follows.

$16^{5} + 16^{4} = 16^{3} (16^{2} + 16) = 16^{2} (272)$

From here we know the the resulting number will be equal to $16^{2} (272)$ now we will need to know the following property. If "b" is divisible by "c" then "ab" (where a is any integer) will also be divisible by c. In our case one of the factors is 272, and it is divisible by 17, so that means that $16^{2} (272)$ will also be divisible by 17.

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A ship is anchored off a long straight shoreline that runs north and south. From two observation points 16 miles apart on shore,

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D - the shortest distance from the ship to the shore.
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3 years ago

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2 years ago

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3 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$