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Ivenika [448]
3 years ago
12

Evaluate the piecewise function at the indicated values from the domain:

Mathematics
1 answer:
lidiya [134]3 years ago
7 0

In this question, we given a piece-wise function, that has different definitions depending on the domain.

Evaluate the function at x = 0.

The exercise asks for us to evaluate the function at x = 0

We have to look at the definition, and see which definition includes x = 0. The equal sign at x = 0 is on the second definition, that is:

f(x) = 1, 0 \leq x < 2

Thus, at x = 0, the value of the function is 1, and the correct answer is given by option A.

For another example of evaluation of a piece-wise function, you can check brainly.com/question/17966003

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1.28

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Use the diagram to complete the statement. Triangle J K L is shown. Angle K J L is a right angle. Angle J K L is 52 degrees and
zzz [600]

Answer:

\bold{sin(38^\circ)=cos(52^\circ)}

Step-by-step explanation:

Given that \triangle KJL is a right angled triangle.

\angle JKL = 52^\circ\\\angle KLJ = 38^\circ

and

\angle KJL = 90^\circ

Kindly refer to the attached image of \triangle KJL in which all the given angles are shown.

To find:

sin(38°) = ?

a) cos(38°)

b) cos(52°)

c) tan(38°)

d) tan(52°)

Solution:

Let us use the trigonometric identities in the given \triangle KJL.

We have to find the value of sin(38°).

We know that sine trigonometric identity is given as:

sin\theta =\dfrac{Perpendicular}{Hypotenuse}

sin(\angle JLK) = \dfrac{JK}{KL}\\OR\\sin(38^\circ) = \dfrac{JK}{KL}....... (1)

Now, let us find out the values of trigonometric functions given in options one by one:

cos\theta =\dfrac{Base}{Hypotenuse}

cos(\angle JLK) = \dfrac{JL}{KL}\\OR\\cos(38^\circ) = \dfrac{JL}{KL}....... (2)

By (1) and (2):

sin(38°) \neq cos(38°).

cos(\angle JKL) = \dfrac{JK}{KL}\\OR\\cos(52^\circ) = \dfrac{JK}{KL} ...... (3)

Comparing equations (1) and (3):

we get the both are same.

\therefore \bold{sin(38^\circ)=cos(52^\circ)}

6 0
3 years ago
Find the product (3a4+4)2​
asambeis [7]

The result is

9

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2

−

16

The reason is the following:

The problem is an example of a notable product: "the sum multiplied by the diference is equal to the difference of squares", that is to say:

(

a

+

b

)

⋅

(

a

−

b

)

=

a

2

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By applying this to our question, we obtain that:

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(

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=

(

3

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−

(

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=

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a

2

−

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