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Kruka [31]
2 years ago
14

Find x. Do not round your answer. 12 cm 6 cm No links

Mathematics
1 answer:
ololo11 [35]2 years ago
8 0

Answer:

679.6

Step-by-step explanation:

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WILL GIVE BRAINLIEST
Paul [167]
256.
EXPLANATION:

When you are using exponents, you are basically just doing repeated multiplication.

Whatever the exponent (smaller number is) is how many times you multiply the base number by itself.

So 4 ^4 is 4 x 4 x 4 x 4. (4 fours.)

So 4 x 4 x 4 x 4 is 256.
HAVE A GUD DAY
7 0
3 years ago
Only one of the comparisons below is correct. Which is correct? What benchmark was used in your answer?
tiny-mole [99]

Answer:

2/3<9/10; I used 3/4 as a benchmark.

Step-by-step explanation:

2/3<1/2; I used 1/2 as a benchmark.

2/3 = 0.(20/3) = 0.667

1/2 = 0.(10/2) = 0.5

So this is wrong, as 0.667 > 0.5.

1/2=3/5; I used 1/4 as a benchmark.

1/2 = 0.(10/2) = 0.5

3/5 = 0.(30/5) = 0.6

0.5 != 0.6, so this is wrong.

2/3<9/10; I used 3/4 as a benchmark.

2/3 = 0.(20/3) = 0.667

9/10 = 0.(90/10) = 0.9

So this is correct, as 0.667 < 0.9

3/4<2/3; I used 1/2 as a benchmark.

3/4 = 0.(30/4) = 0.75

2/3 = 0.(20/3) = 0.667

0.75 > 0.667, so this is wrong.

5 0
3 years ago
Prove that (4n+1)^2(4n-1)^2 is a multiple of 8 for all integers of n.<br><br> thanks.
azamat

Answer: i think  (4n+1)^2(4n-1)^2 isnt a multiple of 8 for all integers of n because:

(4n + 1)²(4n - 1)²

= [(4n + 1)(4n - 1)]²

= (16n² - 1)²

= 16².n².n² - 2.16.n² + 1

= 8n²(32n² - 4) + 1

can see 8n²(32n² - 4) is a multiple of 8 but 1 isnt a multiple of 8

=> (4n + 1)²(4n - 1)² isnt a multiple of 8 for all integers of n.

Step-by-step explanation:

6 0
3 years ago
What is the recursive rule for the sequence?
adoni [48]
An= -2 (an+1), a1 = 5
5 0
3 years ago
Find the invers of f(x)=4/2x+1​
ivanzaharov [21]

Step-by-step explanation:

i hope i have been useful buddy.

good luck ♥️♥️♥️♥️♥️.

8 0
2 years ago
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