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swat32
4 years ago
6

Easy points i just have a question. Is this picture showing up?

Mathematics
2 answers:
dezoksy [38]4 years ago
8 0
Yes the picture is there
tiny-mole [99]4 years ago
6 0
Yes it does have a great day
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Will give brainliest answer
lidiya [134]
The square root of 10 because

sqrt(10) ^2 * 3.14 =

10*3.14 = 31.4
7 0
4 years ago
Read 2 more answers
HELPPPPP
OlgaM077 [116]

Answer:

Point on Midline (0,3)

Maximum (9π/2,3)

Minimum (-9π/2,-3)

Step-by-step explanation:

in the given sine function which is in the form of f(x) = a sin(bx+c) +d

a = amplitude

period = frequency = 18π

Therefore b = 2π/18π = 1/9

Y intercept = vertical shift = 3

Horizontal shift = d = 0

Therefore the sine function will be

f(x) = 6 sin(x/9) + 3

Now first point on the midline is (0,3)

Second point is maximum (9π/2,9)

Third point be a minimum value ( -9π/2,-3)

7 0
3 years ago
What is the value of sec theta given the diagram below?
marishachu [46]

Answer:

\sec \theta=-\sqrt{5}

Step-by-step explanation:

The hypotenuse is h^2=6^2+3^2

h^2=36+9

h^2=45

h=\sqrt{45}

h=3\sqrt{5}

The terminal side of \theta is in the second quadrant.

In this quadrant; the secant ratio is negative.

\sec \theta=-\frac{hypotenuse}{adjacent}

\sec \theta=-\frac{3\sqrt{5}}{3}

\sec \theta=-\sqrt{5}

6 0
3 years ago
Read 2 more answers
What is the equation
AlexFokin [52]

Answer:

y=3x+1

Step-by-step explanation:

Determine slope with two coordinates and use it in the formula

4 0
3 years ago
Of interest is to test the hypothesis that the mean length of all face-to-face meetings and the mean length of all Zoom meetings
Goshia [24]

Answer:

Null hypothesis: \mu_1 = \mu_2 = ..... \mu_j , j =1,2,....,n

Alternative hypothesis: \mu_i \neq \mu_j , i,j =1,2,....,n

The alternative hypothesis for this case is that at least one mean is different from the others.

And the best method for this case is an ANOVA test.

Step-by-step explanation:

For this case we wnat to test if all the mean length of all face-to-face meetings and the mean length of all Zoom meetings are the same. So then the system of hypothesis are:

Null hypothesis: \mu_1 = \mu_2 = ..... \mu_j , j =1,2,....,n

Alternative hypothesis: \mu_i \neq \mu_j , i,j =1,2,....,n

The alternative hypothesis for this case is that at least one mean is different from the others.

And the best method for this case is an ANOVA test.

6 0
3 years ago
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