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3 years ago

10

-2 (x + 3) i need help... do your thing

1 answer:

beks73 [17]3 years ago

8 0

Answer: -3

-2x-6=0

-2x=6

x=-3

Step-by-step explanation:

You might be interested in

How many numbers are equal to the sum of two odd, one-digit numbers?

Dennis_Churaev [7]

Answer:

Seven numbers.

Step-by-step explanation:

Finding the numbers, which are equal to the sum of two odd number and it has to be single digit number.

Lets look into numbers which are odd and single digit.

1 = $1+3, 1+5, 1+7, 1+9$

∴ Sum of the number is $4,6,8\ and\ 10$

3 = $3+5, 3+7, 3+9$

∴ Sum of above number is $8,10\ and\ 12$

5 = $5+7, 5+9$

∴ Sum of above number is $12\ and\ 14$

7= $7+9$

∴ Sum of above number is $16$

Now, accumlating numbers which are fullfiling the criteria, however, making sure no number should get repeated.

∴ Numbers are: $4,6,8,10,12,14\ and\ 16$

Hence, there are total 7 numbers, which are equal to the sum of two odd, one-digit numbers.

4 0

3 years ago

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

1 year ago

-3y-7x <br> When x=6, y=3

Oksi-84 [34.3K]

-9y-42x

Because they are like terms that is the lowest they can go

4 0

2 years ago

Read 2 more answers

Need simplified please

Romashka-Z-Leto [24]

Answer:

70

Step-by-step explanation:

7^2 - ( 6-3^3)

PEMDAS

Parentheses first

7^2 - ( 6-3^3)

The exponent in the parentheses first

7^2 - ( 6-27)

7^2 - ( -21)

Now the exponent

49 - (-21)

Now subtract

49 +21

70

6 0

2 years ago

What's the answer to my answer answer right ok thanks

Anika [276]

Feel free to ask any questions as to what I did

6 0

3 years ago