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creativ13 [48]
3 years ago
5

L An 8.5 quart jug contains how many cups?

Mathematics
1 answer:
pav-90 [236]3 years ago
3 0

Answer:

34 cups

Step-by-step explanation:

just multiply the value in quarts by the conversion factor 4. So, 8.5 quarts times 4 is equal to 34 cups

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What is the product of (3a 2)(4a2 – 2a 9)?12a3 − 2a 1812a3 6a 912a3 − 6a2 23a 1812a3 2a2 23a 18
timama [110]
To get the product of (3a + 2) and (4a² - 2a + 9), we can apply the rules of distribution over multiplication and have 3a(4a² - 2a + 9) + 2(4a² - 2a + 9) 12a³ - 6a² + 27a + 8a² - 4a + 18 Adding like terms of the same degree, we can simplify it into 12a³ + 2a² + 23a + 18 Therefore, the answer is D<span>. </span>
5 0
3 years ago
Read 2 more answers
The length of a box as reported by its manufacturer is 13.67 cm. Joe has a ruler which has markings for each millimeter. Which m
AlekseyPX

Answer:

Option (3)

Step-by-step explanation:

Joe has a ruler which has markings for each millimeter, so the least measurement which Joe can do is in millimeter.

Since, 10 mm = 1 cm

Therefore, 1 mm = 0.1 cm

This rule states that Joe can measure a length or distance in nearest tenth of a cm only.

If length of a box = 13.67 cm,

So Joe can measure it as 13.7 cm approximately.

Therefore, Option (3) will be the answer.

6 0
3 years ago
One number is seven less than two times another. If their sum is decreased by four, the result is one. Find the numbers.
allsm [11]

Answer:

(2n + k) - 7

Step-by-step explanation:

1. Let ''a number'' be presented as n

2. ''twice a number'' → 2 n

3. ''twice a number and k'' → 2n + k

4. 7 less than twice a number → (2h + k) - 7

Google it!

5 0
3 years ago
Can someone please help and please show work
pychu [463]

Answer:


Step-by-step explanation:

4 and a half

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
1 year ago
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