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algol [13]
3 years ago
5

True or False: The value of r represents the reference angle when plotting a point in polar coordinates.

Mathematics
1 answer:
timama [110]3 years ago
4 0
<span>The value of r represents the reference angle when plotting a point in polar coordinates.
False
</span>
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i have 6 digits. One of my 3s is worth 300.000. The other is worth 1/10 as much. My 6 is worth 600.The rest of my digits are zer
motikmotik
The way you wrote the problem makes the answer 330,600. I think your decimals were meant to be commas. If I am wrong, please forgive me.
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3 years ago
I need helpppppppppppp
ratelena [41]
The answer to question 3 is D. 15(4a+3). And the answer to question 4 is A. 90. Have a great day
7 0
2 years ago
In response to a gss question in 2006 about the number of hours spent per day watching television, the responses by the fifteen
NemiM [27]

Answer:

Standard error = 0.4

Step-by-step explanation:

Step 1

We find the Standard Deviation

The formula = √(x - mean)/n - 1

n = 15

Mean = 1.93 hours

= √(0- 1.93)² + (0-1.93)² +(0- 1.93)²+( 0- 1.93)²+ (1- 1.93)² + (1- 1.93)² +(1 - 1.93)² +(2 - 1.93)² + (2 - 1.93)² + (2 - 1.93)² + (2 - 1.93)² + ( 2 - 1.93)² +(4 - 1.93)² +(4 - 1.93)² + (5 - 1.93)²/15 - 1

= √(3.737777776 + 3.737777776 + 3.737777776 + 0.871111111 +0.871111111 + 0.871111111 + 0.004444444445+ 0.004444444445 + 0.004444444445 + 0.004444444445 + 0.004444444445 + 1.137777778 + 4.271111112 + 4.271111112 + 9.404444446)/15 - 1

= √2.352380952

= 1.533747356

Step 2

We find the standard error

The formula = Standard Deviation/√n

Standard deviation = 1.533747356

n = 15

= 1.533747356/√15

= 1.533747356 /3.87298334621

= 0.39601186447

Approximately = 0.4

Therefore, the standard error is 0.4

4 0
2 years ago
The integral represents the volume of a solid. describe the solid. π π 0 sin(x) dx
Oxana [17]
Is that all the question says ??
4 0
3 years ago
Determine whether the statement is true or false, and explain why.
Zielflug [23.3K]

Answer:

False

Step-by-step explanation:

In this case, i woudnt reccomend to use integration by parts, becuase you are not simplifying the expression by doing an integration and a derivation. It is not easy to integrate 1/x³+1, and if you derivate it, then a natural logarithm would appear, and the integral wont be easier after this parts step.

It is a better idea to use integration by substitution. Note that if you replace x³+1 by a variable y, we have that dy = 3x² dx. We can easily make a 3xdx appear in the integral by multiplying and dividing by 3, solving the integral easily:

\int\limits_0^1 x^2/x^3+1 \, dx = 1/3 * \int\limits_1^2 du/u = (ln(2) - ln(1) )/3

(Note that, if x ranges from 0 to 1, then u = x³+1 ranges from 0³+1 = 1 to 1³+1 = 2)

4 0
2 years ago
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