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skad [1K]
4 years ago
13

Is the square root of infinity still infinity?

Mathematics
1 answer:
Jet001 [13]4 years ago
6 0
Infinity isn't actually a number - it's a concept that represents that the set of numbers goes on forever.  As a result, you can't take the square root of infinity.  However, you could state that as x approaches infinity (or rather, increase indefinitely), the square root of x also approaches infinity (again, increases indefinitely).
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A wheel is shaped like a circle with a diameter of 35 cm. To the nearest
zhenek [66]

Answer:

C = 110 cm

Step-by-step explanation:

The circumference is given by

C = pi *d

C = 3.14 * d

C = 3.14 * 35

C =109.9

Rounding

C = 110 cm

3 0
4 years ago
Section 1 - Question 1
xenn [34]

Value of 5(d–2) is 45.

Option C is correct.

Step-by-step explanation:

We are given 3(d+1)=4(d–2) and we need to find the value of 5(d–2)

First we will solve 3(d+1)=4(d–2)  to find value of d:

Solving:

3(d+1)=4(d–2) \\3d+3=4d-8\\3d-4d=-8-3\\-d=-11\\d=11

So, value of d = 11

Now putting value of d in 5(d–2) and finding its value:

5(d–2)\\=5(11-2)\\=5(9)\\=45

So, value of 5(d–2) is 45.

Option C is correct.

Keywords: Solving Equations

Learn more about Solving Equations at:

  • brainly.com/question/1563227
  • brainly.com/question/2403985
  • brainly.com/question/11229113

#learnwithBrainly

6 0
4 years ago
What is 600.00 lb into g?
ollegr [7]

272155

I hope I helped, good luck

4 0
3 years ago
Solve A = h (a + b) for b.
fenix001 [56]

Answer:

b=A/h-a

Step-by-step explanation:

A=h(a+b)

a+b=A/h

b=A/h-a

8 0
3 years ago
choose a number between 67 and 113 that is a multiple of 4, 8, and 16 . Write all the numbers that she could choose. If there is
Allushta [10]
2 is a multiple of 4, so anything that's a multiple of 4 will automatically be a multiple of 2 so 3 x 4 = 12 so multiples of 12 that are between 67 and 113 are...? 12 x 6 = 72 12 x 7 = 84 12 x 8 = 96 12 x 9 = 108 <span> 72 or 84 or96 or 108 all are divisible by 2,3 and 4 all fall between 67 and 112</span>
6 0
4 years ago
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