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Juli2301 [7.4K]
1 year ago
15

A. How many feet is

5}" align="absmiddle" class="latex-formula"> of a mile? ______
b. How many feet is \frac{1}{100} of a mile? ______
Mathematics
1 answer:
kipiarov [429]1 year ago
8 0

Answer:

a. 1056 ft

b. 52.8 ft

Step-by-step explanation:

1 mile = 5280 feet

a.

\frac{1}{5} *5280 = 1056

Therefore 1/5 of a mile is 1056 feet.

b.

\frac{1}{100} *5280=52.8

Therefore 1/100 of a mile is 52.8 feet

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Every integer number is also a. Irrational number b. Whole number c. Natural number d. Rational number
const2013 [10]

Answer:

Option d

Step-by-step explanation:

Other options are not eligible because

a) Integers are not irrational

b)Whole numbers start from 0 and consists positive integers only.But integers consist negative integers also.

c)Natural number consists of positive integers only.

If the question had been asked as some integers are also, then options b) and c) could have been written . But in this case , it is asked every integer is also.

Thank you!

5 0
2 years ago
Which graph shows a proportional relationship?
stiks02 [169]
C and D because in that image that is what an proportional relationship looks like so anything that looks like the graph in the photo or c and d is proportional

6 0
2 years ago
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Explain the steps that are necessary to solve 4x - 2(x + 3) = x + 1
Sidana [21]
So you would distribute what's in the parenthesis and end up with 4x-2x-6=x+1, then you combine like terms and end up with 2x-6=x+1, now you can add 6 to both sides or subtract 1 both sides (basically anything goes for this step just make sure you apply the same thing on both sides) so I'll add 6 and continue just to show you... then you end up with 2x=x+7 then you subtract 1x on both sides to get x=7. sorry it's a lot hopefully you get it know!
8 0
3 years ago
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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
1 year ago
Please help me my teacher won’t help me at all :/
Rudik [331]

Answer:

1 to 75

2 to 120

3 to 165

4 to 210

5 to 255

Step-by-step explanation:

5 0
2 years ago
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