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katrin2010 [14]
3 years ago
5

What is the greatest fraction you can make using the digits 4, 7, and 9?

Mathematics
1 answer:
wel3 years ago
6 0

Answer:

  (7^9)/4 = 40,353,607/4

Step-by-step explanation:

Assuming each digit is used once and exponentiation is allowed, the largest numerator and smallest denominator will result in the largest fraction.

__

If other functions, such as factorial are allowed, then there might need to be a limit on the number of times they are applied. For example,

  (7!)^(9!)/4 has about 1 million digits

something like ...

  ((7!)^(9!))!/4 has many more digits than that

and you can keep piling on the factorial symbols to any desired depth.

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A certain recipe asks for 4/9 cup of granola , How much is needed to make 4/9 of a recipe
artcher [175]
4/9 times 4/9 is 16/81.
To make 4/9 of a recipe, 16/81 cups of granola needed.
7 0
3 years ago
Noah planned a parallelogram-shaped sidewalk to lead from the street to the front door of his house. He realized the walkway was
nekit [7.7K]

Answer:

Step-by-step explanation:

8 0
3 years ago
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What is the distance between -7.5 and -15.3 on a number line
timama [110]
Okay. To find the distance apart from each other, subtract 15.3 and 7.5. When you do that, you should get 7.8. -7.5 and -15.3 are 7.8 units away on a number line.
8 0
3 years ago
Which of the following has a solution set of {x | x = 0}?
notka56 [123]

Answer:

(b)  (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) =  {x | x = 0}

Step-by-step explanation:

Here, the given expression is : {x | x = 0}

So, the ONLY element in the given set = {0}

Now, take each option and solve the given expression:

(a)  x + 1 < -1

Adding -1 BOTH sides, we get:

x + 1 -1 < -1  -1

or, x < - 2 ⇒ x = { -∞ , .... , -4,-3}

Also,   x + 1 < 1

Adding -1 BOTH sides, we get:

x + 1 -1 < 1  -1

or, x <0 ⇒ x = { -∞ , .... , -4,-3,-2,-1}

So, (x + 1 < -1) ∩ (x + 1 < 1) =  { -∞ , .... , -4,-3}∩ { -∞ , .... , -4,-3,-2,-1}  

=  { -∞ , .... , -4,-3}

⇒ (x + 1 < -1) ∩ (x + 1 < 1) ≠ {0}

Similarly, solving

(b) (x + 1 ≤ 1) ∩ (x + 1 ≥ 1)

(x + 1 ≤ 1) =  x≤ 0 =   { -∞ , .... , -4,-3,-2,-1, 0}

(x + 1 ≥ 1) =  x ≥ 0 =  {0,1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = {0}

(b)(x + 1 < 1) ∩ (x + 1 > 1)

(x + 1 < 1) =  x <  0 =   { -∞ , .... , -4,-3,-2,-1}

(x + 1 > 1) =  x  > 0 =  { 1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = Ф

Hence,  (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) =  {x | x = 0}

8 0
4 years ago
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Solve. Please show your work
Sphinxa [80]

Answer:

128

Step-by-step explanation:

8^2√2^2

64√2^2              rewrite √2^2 as 2

64*2

=128

5 0
3 years ago
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