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frosja888 [35]
3 years ago
13

Show that cos^2 330degrees=1

Mathematics
1 answer:
Sholpan [36]3 years ago
4 0
Actually it's not 1 

330 in 4th quadratic is just 30 in 1st and cos is +ve

so you have

sqrt(3)/2 ^2 = 3/4 = 0.75 
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Three randomly selected households are surveyed. /The numbers of people in the households are 1, 3, and 8. Assume that samples o
inn [45]
The question is asking to find the variance for the said samples in the problem ans use the sample data to determine each variance, and base on my further computation and further calculation, I would say that the answer would be the following:
#1. 3.3 -> 1 and 3 -> 2/9
#2. 1-> 0->3/9
#3. 6.3 - > 8 and 3-> 2/9
#4 49-> 1 and 8-> 2/9
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2 years ago
Three friends order a pizza. Danny eats 1/2 of the pizza, Mike eats 1/3 of the pizza, and Chris eats 1/6 of the pizza. The next
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It Mike who ate the most pizza

Step-by-step explanation:

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3 years ago
I need helpASAP!! Which point would map onto itself after a reflection across the line y = –x?
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3 years ago
Given f(x)= 7x-5. Find the value of f(8)​
e-lub [12.9K]

Answer: 52

Explanation:

You need to replace the x-values with 8 to solve this. So, the equation would look like this. f(8)= 7 x 8 - 5

7 times 8 is 56. 56 minus four is 52 which is your answer.

If you needed to graph this then it would look like this... (8, 52)

The 8 is in the input because it is the x-value (we replace x with 8) and 52 is output/range/y-value because it is the answer.

8 0
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PLEASE HELP QUICK!! WILL OFFER 100 POINTS TO THE FIRST ONE THAT ANSWERS
Reil [10]

Given

x+1 = \sqrt{7x+15}

We have to set the restraint

x+1\geq 0 \iff x \geq -1

because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.

Similarly, we have

x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}

So, we have to impose

x-7\geq 0 \iff x \geq 7

Squaring both sides, we have

(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.

7 0
3 years ago
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