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3 years ago

Divide 30 by it's half, add the half of 18 that has been divided by the half of the half of 12 after been divided by half.

Mathematics

1 answer:

WARRIOR [948]3 years ago

6 0

Answer:

We want to divide 30 by it's half.

The half of 30 is 30/2 = 15

then we have: 30/15 = 2

Now we want to add the half of 18, that has ben divided by the half of the half of 12 hafter been divided by half.

The half of 12 is 12/2 = 6

the half of the half of 12 is 6/2 = 3

now, we wanto to divide the half of 18 by that number.

The half of 18 is 18/2 = 9

then we have 9/3 = 3

and we wanted to add this to the number previous found:

2 + 3 = 5

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Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$

Viktor [21]

When $n=0$ , we have

$5^{3^0} + 1 = 5^1 + 1 = 6$

$3^{0 + 1} = 3^1 = 3$

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for $n=k$ , that

$3^{k + 1} \mid 5^{3^k} + 1$

Now for $n=k+1$ , we have

$5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)$

so we know the left side is at least divisible by $3^{k+1}$ by our assumption.

It remains to show that

$3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1$

which is easily done with Fermat's little theorem. It says

$a^p \equiv a \pmod p$

where $p$ is prime and $a$ is any integer. Then for any positive integer $x$ ,

$5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3$

Furthermore,

$5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3$

which goes all the way down to

$5^{3^k} \equiv 5 \pmod 3$

So, we find that

$\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3$

QED

5 0

2 years ago

Please please help me

Anarel [89]

so you are going to use the Pythagorean theorem on the smaller triangle to find the connecting side's length

${5}^{2} = {3}^{2} ( {b}^{2} ) \\ b = 4$

now you can find the bigger triangle's hypotenuse

${6}^{2} + {4}^{2} = {c}^{2} \\$

c=4

8 0

3 years ago

There are 515 students who attend central middle school. Three out of every 5 students live within 1 mile of the school. How man

AleksandrR [38]

Answer:

103 students live 1 mile away from the school.

Step-by-step explanation:

First dicide what the problem is really asking you then you end up getting divisoin them divide 515/3.

5 0

3 years ago

Given AB=10cm, CD=11cm, and AD=39cm, find the length of BC

liraira [26]

Answer:

BC = 18 cm

Step-by-step explanation:

As we can see from the diagram:

AB + BC + CD = AD

⇒ 10 cm + BC + 11 cm = 39 cm

⇒ BC + 21 cm = 39 cm

⇒ BC = 39 cm - 21 cm

⇒ BC = 18 cm

7 0

2 years ago

8. f(n) = f(n-1) + 1; f(1) = win

DanielleElmas [232]

Answer:

we are looking for F

but in the question it stated that f(n) and at the end it also stated that f(1) so n=1

Step-by-step explanation:

we are using BODMAS

f(n-1)+1

f(1-1)+1=0+1

f=1

6 0

3 years ago