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VikaD [51]
3 years ago
14

I need help in this before 11:59 please

Mathematics
1 answer:
Aliun [14]3 years ago
4 0

Step-by-step explanation:

√7 * √7 = 7 , which is not a irrational number

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Easy Points<br><br> whats 1=1
Katen [24]

infinite set is the answer

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3 years ago
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A restaurant manager collected data on the number of customers in a party in the restaurant and the time elapsed until the party
Stella [2.4K]

Answer:

Option C The parties with a larger number of customers are associated with the longer times elapsed until the party left the restaurant.

Step-by-step explanation:

The correlation coefficient 0.78 shows that positive association between two variables number of customers and elapsed time until party left restaurant.

The positive association means that as the number of customers in a party increases the elapsed time also increase. So, we can say that the parties with a larger number of customers are associated with the longer times elapsed until the party left the restaurant.

4 0
3 years ago
what is the value of 8/15÷(-0.35) A -75/14 B -32/21 C -21/32 D -14/75 I would really appreciate it if you could also explain how
barxatty [35]
Answer;-1.5238095238
First step; 8/15

8/15 = 0.53 and so on in a repeating form. (0.53333333333)

Divide; 0.53 / -0.35 

= -1.5238095238

5 0
3 years ago
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If the actual length of a boat is 120 inches and the model of the same boat is 6 inches, identify the scale factor of these two
docker41 [41]

Answer:

1/20

Step-by-step explanation:

4 0
3 years ago
Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A
iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA

7 0
3 years ago
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