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Maru [420]
3 years ago
12

I need help with this question,

Mathematics
1 answer:
Paraphin [41]3 years ago
3 0
The slope of the line would be 3/4
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Use complete sentences to describe the domain of the consine function.
Anika [276]
Can you provide a picture ?? or some sort of context
3 0
3 years ago
Read 2 more answers
Use the given minimum and maximum data​ entries, and the number of​ classes, to find the class​ width, the lower class​ limits,
Gennadij [26K]

Using proportions and the information given, it is found that:

  • The class width is of 14.375.
  • The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.
  • The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.

-------------------------

  • Minimum value is 19.
  • Maximum value is of 134.
  • There are 8 classes.
  • The classes are all of equal width, thus the width is of:

W = \frac{134 - 19}{8} = 14.375

-------------------------

The intervals will be of:

19 - 33.375

33.375 - 47.750

47.750 - 62.125

62.125 - 76.500

76.500 - 90.875

90.875 - 105.250

105.250 - 119.625

119.625 - 134.

  • The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.
  • The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.

A similar problem is given at brainly.com/question/16631975

6 0
3 years ago
Can someone pls help! 20 points!
Aleksandr-060686 [28]
J=3,1 and 0,5 K=1,0 and 0,-2 L=1,3 and 0 M=3,-3 and 0,-2 N=4,6 and 0,5
6 0
3 years ago
Read 2 more answers
Assume that the terminal side of thetaθ passes through the point (negative 12 comma 5 )(−12,5) and find the values of trigonomet
zmey [24]

Answer:

\sin \theta = \dfrac{5}{13} and \sec \theta = -\dfrac{13}{12}

Step-by-step explanation:

Assume that the terminal side of thetaθ passes through the point (−12,5).

In ordered pair (-12,5), x-intercept is negative and y-intercept is positive. It means the point lies in 2nd quadrant.

Using Pythagoras theorem:

hypotenuse^2=perpendicular^2+base^2

hypotenuse^2=(5)^2+(12)^2

hypotenuse^2=25+144

hypotenuse^2=169

Taking square root on both sides.

hypotenuse=13

In a right angled triangle

\sin \theta = \dfrac{opposite}{hypotenuse}

\sin \theta = \dfrac{5}{13}

\sec \theta = \dfrac{hypotenuse}{adjacent}

\sec \theta = \dfrac{13}{12}

In second quadrant only sine and cosecant are positive.

\sin \theta = \dfrac{5}{13} and \sec \theta = -\dfrac{13}{12}

6 0
3 years ago
The numerical coefficient of -2pqr
VMariaS [17]

Answer:  -2

Explanation: The coefficient is the number to the left of the variable.

Another example would be that 10xy has the coefficient 10.

Something like x^2 has coefficient 1 because x^2 = 1x^2.

7 0
3 years ago
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