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victus00 [196]
3 years ago
7

I need help anyone??? please??????

Mathematics
2 answers:
olchik [2.2K]3 years ago
8 0
The principal square root of 49 is just 7 because 7 times 7 is 49.
It can not be -7 because a principle square root cannot be negative.

\sf\ Answer: A)\ 7

\sf\ \sqrt{49}=\\\±7\\\\-7\ and\ 7\ are\ both\ square\ roots\ of\ 49\ because\ 7\times7=49\ and\\\\-7\times-7=49.\\\\But\ as\ I\ said\ before,\ principal\ square\ roots\ can\ not\ be\ negative\\\\numbers.\\\\That\ will\ eliminate\ any\ options\ with\ -7.\\\\So\ the\ only\ principal\ square\ root\ that\ we\ have\ is\ {\boxed{7}
Ronch [10]3 years ago
5 0
The answer would be 7 
because the square root of 49 is 7. 
so 7 multiplied 7 is ..49

The answer wont be -7 because in Principal root, 
negative numbers are not involved, you cant do it! :)

That being said you'd also be able to know that since no negative numbers are allowed in principle root, -7 and 7 are out as choices, and D is totally wrong since we now know what the priniciple root means :)
The answer is 7
Hope this helped !

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2 years ago
Read 2 more answers
F(x) = (x - 4)(x + 5)
drek231 [11]

Answer:

x  

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Step-by-step explanation:

6 0
3 years ago
4. If you spin the spinner twice, what is the probability that the spinner will land on 5 on the first spin and 2 on the second
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Answer:

1/16

Step-by-step explanation:

Given in comment:

5,4,3,2 = Spinner

Thus, there are 4 possible answer

Given:

If you spin the spinner twice, what is the probability that the spinner will land on 5 on the first spin and 2 on the second spin?

1/16

3/16

1/8

1/4

Solve:

Knowing that there only <u>One 5 and One 2 on the spinner.</u>

Thus, in fraction form - 1/4 and 1/4

Probability land on 5 = 1/4

Probability land on 2 = 1/4

Because there are 4 possible solution and one of each is 5,4,3,2

Now, multiply then together:

1/4 × 1/4 = {1 ×1 = 1 and 4 × 4 = 16}

Put Together:

1/16

Hence, the answer is " The Probability that the spinner will land on 5 on the first spin and 2 on the second spin is 1/16  

<em>~Lenvy~</em>

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