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3 years ago

Which of the following equations matches the function shown above?

Mathematics

1 answer:

yanalaym [24]3 years ago

5 0

Answer:

C=y=sin1/2x

Step-by-step explanation:

As given in the graph:

Amplitude= 1

period=2π

Finding function of sin that have period of 4π and amplitude 1

A: y=1/2sinx

Using the formula asin(bx-c)+d to find the amplitude and period

a=1/2

b=1

c=0

d=0

Amplitude=|a|

=1/2

Period= 2π/b

=2π

B: y=sin2x

Using the formula asin(bx-c)+d to find the amplitude and period

a=1

b=2

c=0

d=0

Amplitude=|a|

Period= 2π/2

=π

C: y=sin1/2x

Using the formula asin(bx-c)+d to find the amplitude and period

a=1

b=1/2

c=0

d=0

Amplitude=|a|

Period= 2π/1/2

=4π

D: y=sin1/4x

Using the formula asin(bx-c)+d to find the amplitude and period

a=1

b=1/4

c=0

d=0

Amplitude=|a|

Period= 2π/1/4

=8π

Hence only c: y=sin1/2x has period of 2π and amplitude 1

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Geometry help please!

photoshop1234 [79]

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Hope this helps!! :)

8 0

3 years ago

Brahmagupta’s solution to a quadratic equation of the form ax2 + bx = c involved only one solution. Which solution would he have

zlopas [31]

Answer:

0.897

Step-by-step explanation:

Brahmagupta formula for quadratic equation $ax^2+bx=c$ is

$x=\dfrac{\sqrt{4ac+b^2}-b}{2a}$

It involved only one solution.

The given equation is

$3x^2+4x=6$

Here, a=3, b=4 and c=6. Put these values in the above formula.

$x=\dfrac{\sqrt{4(3)(6)+(4)^2}-4}{2(3)}$

$x=\dfrac{\sqrt{72+16}-4}{6}$

$x=\dfrac{\sqrt{88}-4}{6}$

$x\approx \dfrac{5.38}{6}$

$x\approx 0.897$

Therefore, the required solution is 0.897.

7 0

3 years ago

Write a proportion to find how many points x a student needs to score on a test worth 50 points to get a test score of 40%. $$On

lesya [120]

2 ways to describe the proportion of 40% out of 50 points are:

20/50
or
2:5

7 0

4 years ago

Please help me…………….

valina [46]

9514 1404 393

Answer:

54.8 km

Step-by-step explanation:

The sketch and the applicable trig laws cannot be completed until we understand what the question is.

Given:

two boats travel for 3 hours at constant speeds of 22 and 29 km/h from a common point, their straight-line paths separated by an angle of 39°

Find:

the distance between the boats after 3 hours, to the nearest 10th km

Solution:

A diagram of the scenario is attached. The number next to each line is the distance it represents in km.

The distance (c) from B1 to B2 can be found using the law of cosines. We can use the formula ...

c² = a² +b² -2ab·cos(C)

where 'a' and 'b' are the distances from the dock to boat 1 and boat 2, respectively, and C is the angle between their paths as measured at the dock.

The distance of each boat from the dock is its speed in km/h multiplied by the travel time, 3 h.

c² = 66² +87² -2·66·87·cos(39°) ≈ 3000.2558

c ≈ √3000.2558 ≈ 54.77

The boats are about 54.8 km apart after 3 hours.

7 0

3 years ago

What is the value of x ??

RideAnS [48]

Answer:

4.2mm

Step-by-step explanation:

This are two similar triangles.

Comparing the two:

8/3 = (7+x)/x

Cross multiplying

8x = 3 (7+x)

8x = 21 + 3x

8x - 3x = 21

5x = 21

Dividing by 5

X = 21/5

X = 4.2mm

3 0

3 years ago