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ale4655 [162]
3 years ago
6

Pls help I never did this before

Mathematics
2 answers:
Kipish [7]3 years ago
7 0

Answer:

\sqrt{16} and \sqrt[3]{64}

Step-by-step explanation:

let me know if u need this explained

maria [59]3 years ago
6 0

Answer:

16 and 64

Step-by-step explanation:

When you sqaure root 16, you get 4

When you cube root 64, you get 4

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there are 18 students ob the school bus. there are 5 seniors 2 juniors,6 sophomores and 5 freshmen. what is the ratio of juniors
dimaraw [331]
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6 0
4 years ago
If p=m and n=16 in the equation m/45 = n/p, what is the value of m?
Andrej [43]
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6 0
3 years ago
The sum of 3 consecutive odd numbers is 177. What is the value of the smallest number in the sequence
riadik2000 [5.3K]

Ok this is a very question that can solved very easily. So we know that there three numbers, and they are consecutive, this means that they are all one bigger than the number before so for example 1,2, and 3 or 20, 21, and 22. These numbers are called consecutive. Now let's just pretend that the answer to this question is the variable "x". Now x is the smallest number in all the three numbers.

Using our knowledge of consecutive numbers we can say that the next to numbers will be one more than the number before so:

(x+1) and (x+2)

So we have our three numbers: x , x+1 , and x+2. Now let's set these numbers equal to 177.

177 = x + x + 1 + x + 2 → Combine like terms.

177 = 3x + 3 → Subtract 3 from both sides.

174 = 3x → Divide both sides by 3

58 = x

So that means that our first and smallest number is going to be 58 our second number will be +1 so 59 and our third number will be +2 so 60.

So the three numbers will be 58, 59, and 60.

Hope this helped.

4 0
3 years ago
1.What is the intersection of all the open intervals containing the closed interval [0,1]? Justify your answer.
ExtremeBDS [4]

Answer:

Step-by-step explanation:

1)

We can find an open interval (-\frac{1}{n}, 1+\frac{1}{n}). Which contains the closed interval [0,1] when n \to \infinity

Therefore, the intersection of all the open intervals containing [0,1[/tex[ is [tex][0,1]

2)

We can find closed intervals containing [0,1] in (-\frac{1}{n}, 1+\frac{1}{n}) when n \to \infinity

Since (-\frac{1}{n}, 1+\frac{1}{n}) is closed, the intersection of all those intervals containing [0,1[/tex[ is  the closed interval [tex][0,1]

7 0
3 years ago
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