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Vedmedyk [2.9K]
3 years ago
13

Is it just me, or does the my pillow ad play on every other break on tv

Mathematics
1 answer:
IRINA_888 [86]3 years ago
3 0

Answer:

Yes..

Step-by-step explanation:

every old TV channel you see a glitchy my pillow ad

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Find the gcf of <br> 24, 14, 21
kobusy [5.1K]

Answer:

1

Step-by-step explanation:

All the numbers provided can have a factor of 1.

eg.)

1*24=24

1*14=14

1*21=21

This is because there is only one factor they all commonly share, 1.

4 0
4 years ago
Read 2 more answers
What is the solution to 2x+12=5x-9
Wittaler [7]
So 2x+12=5x-9
you want all of the unknowns on one side and all of  the known values on the other or
subtract 2x from both sides
12=3x-9
add 9 to both sides
21=3x
divide both sides by 3
7=x
4 0
3 years ago
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Which expression is equivalent to 4 (9 + 7)
Bad White [126]

4(9+7) = 36+28 = 64

open the parentheses, then simplify.

8 0
2 years ago
Save Work out the size of angle AED.<br> b) Work out x
Mamont248 [21]

Answer:

x = 10°

Step-by-step explanation:

a). Since, opposite angles of a cyclic quadrilateral are supplementary angles"

   Therefore, in cyclic quadrilateral ABDE,

   m∠ABD + m∠AED = 180°

   110° + m∠AED = 180°

   m∠AED = 180° - 110°

                 = 70°

b). AD = ED [Given]

   m∠EAD = m∠AED [Since, opposite angles of equal sides are equal in measure]

   m∠EAD = m∠AED = 70°

   By triangle sum theorem in ΔABD,

   m∠BAD + m∠ABD + m∠ADB = 180°

   m∠BAD + 110° + 40° = 180°

   m∠BAD = 180 - 150

                 = 30°

    m∠AEB = m∠AED + m∠DAB [By angles addition postulate]

    m∠AEB = 70° + 30°          

                  = 100°

    By triangle sum theorem in the large triangle,

    x° + m∠AEB + m∠EAB = 180°

    x° + 100° + 70° = 180°

    x = 180 - 170

    x = 10°

8 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
2 years ago
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