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Marizza181 [45]
3 years ago
13

Find the measure of the are or angle indicated. Assume that lines which appear tanget are tangents. ​

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
8 0

9514 1404 393

Answer:

  1. 85°
  2. 60°

Step-by-step explanation:

1. Angle JKL is half the measure of the intercepted arc JK.

  (1/2)JK = 1/2(360° -190°) = (1/2)(170°) = 85°

  angle JKL is 85°

__

2. The angle between tangents is the supplement of the intercepted arc.

  angle JKL = 180° -(360° -240°)

  angle JKL = 60°

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Base: z(x)=cosx Period:180 Maximum:5 Minimum: -4 What are the transformation? Domain and Range? Graph?
garik1379 [7]

Answer:

The transformations needed to obtain the new function are horizontal scaling, vertical scaling and vertical translation. The resultant function is z'(x) = \frac{1}{2}  + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right).

The domain of the function is all real numbers and its range is between -4 and 5.

The graph is enclosed below as attachment.

Step-by-step explanation:

Let be z (x) = \cos x the base formula, where x is measured in sexagesimal degrees. This expression must be transformed by using the following data:

T = 180^{\circ} (Period)

z_{min} = -4 (Minimum)

z_{max} = 5 (Maximum)

The cosine function is a periodic bounded function that lies between -1 and 1, that is, twice the unit amplitude, and periodicity of 2\pi radians. In addition, the following considerations must be taken into account for transformations:

1) x must be replaced by \frac{2\pi\cdot x}{180^{\circ}}. (Horizontal scaling)

2) The cosine function must be multiplied by a new amplitude (Vertical scaling), which is:

\Delta z = \frac{z_{max}-z_{min}}{2}

\Delta z = \frac{5+4}{2}

\Delta z = \frac{9}{2}

3) Midpoint value must be changed from zero to the midpoint between new minimum and maximum. (Vertical translation)

z_{m} = \frac{z_{min}+z_{max}}{2}

z_{m} = \frac{1}{2}

The new function is:

z'(x) = z_{m} + \Delta z\cdot \cos \left(\frac{2\pi\cdot x}{T} \right)

Given that z_{m} = \frac{1}{2}, \Delta z = \frac{9}{2} and T = 180^{\circ}, the outcome is:

z'(x) = \frac{1}{2}  + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right)

The domain of the function is all real numbers and its range is between -4 and 5. The graph is enclosed below as attachment.

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3 years ago
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Answer:

f(-3)=7

Step-by-step explanation:

just plug in -3

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