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3 years ago

What is the equation for this graph?

Mathematics

1 answer:

Sonja [21]3 years ago

8 0

Answer:

y = -2 sin 6x.

Step-by-step explanation:

The parent graph sin x has been reflected in the x -axis because the graph descends from the origin. Also the graph has been compressed horizontally by a factor 6. The amplitude has been doubled so it must have a 2 before the sine.

The equation of the graph is -2sin6x

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Approximate the real zeros of f(x) = x2 + 3x + 1 to the nearest tenth

pshichka [43]

Approximate the real zeros of f(x) = x2 + 3x + 1 to the nearest tenth

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3 years ago

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What is greater 8ft or 3yd

alexdok [17]

3yds because 1 yards is 3 feet so 3 times 3=9 feet greater than 8feet

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3 years ago

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How many solutions does the system of equations below have? (PLEASE HELP)

Artist 52 [7]

Answer:

y= -54-2

y= -56 &

x= -54/ -9

x= 6

Step-by-step explanation:

y= -9x-2

-9x= y+2 divide bothe side by -9

x=y+2/-9

y= -9(y+2/ -9) -2 = (y+2)-2

y= -2(y+2)

y= -2y-4

y+2y= -4

3y= -4 divide bothe side by 3

y= -4/3

x=y+2/ -9

x= -4/3+2/-9

x= -2/3× -9

x=18/3

x=6

6 0

3 years ago

The height of a tennis ball tossed into the air is modeled by h(x) = 40x â€" 16x 2, where x is elapsed time in seconds. During w

egoroff_w [7]

The time interval x >0.5 and x > 2.04 will the tennis ball be at least 15 feet above the ground.

Given that,

The height of a tennis ball tossed into the air is modeled by,

$\rm h(x) = 40x - 16x^2$

Where x is elapsed time in seconds.

We have to determine,

During what time interval will the tennis ball be at least 15 feet above the ground?

According to the question,

The height of a tennis ball tossed into the air is modeled by,

$\rm h(x) = 40x - 16x^2$

Where x is elapsed time in seconds.

Then,

When the tennis ball be at least 15 feet above the ground,

$h(x) = 15$

Substitute the value of h(x) in the equation,

$\rm h(x) = 40x - 16x^2\\\\ 15 = 40x - 16x^2\\\\ 16x^2-40x+15=0$

Factorize the equation for finding the time interval will the tennis ball be at least 15 feet above the ground is,

$\rm 16x^2-40x+15= 0\\\\x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\x = \dfrac{-(-40)\pm \sqrt{(-40)^2-4 \times 16 \times 15}}{2 \times 16}\\\\x = \dfrac{-(-40)\pm \sqrt{1600- 960}}{32}\\\\x = \dfrac{40\pm \sqrt{640}}{32}\\\\x = \dfrac{40 + 25.29}{32} \\\\x = \dfrac{40 + 25.29}{32} \ and \ x = \dfrac{40 - 25.29}{32}\\\\x = \dfrac{65.29}{32} \ and \ x = \dfrac{14.71}{32}\\\\x = 2.04 \ and \ x = 0.45$

Hence, The time interval x >0.5 and x > 2.04 will the tennis ball be at least 15 feet above the ground.

For more details about Inequality refer to the link given below.

brainly.com/question/17177510

7 0

3 years ago

Find all angles, 0≤θ<360, that satisfy the equation below, to the nearest 10th of a degree. 4tan^2 θ+tanθ=−4tanθ−1

Serga [27]

Answer:

135 and 315degrees

Step-by-step explanation:

Given the expression

4tan²θ+tanθ=−4tanθ−1

Equate to zero

4tan²θ+tanθ+4tanθ+1 = 0

4tan²θ+5tanθ+1 = 0

Let x = tanθ

4x²+5x+1 = 0

Factorize;

4x²+4x+x+1 = 0

4x(x+1)+1(x+1) = 0

4x+1 = 0 and x+1 = 0

4x = -1 and x = -1

x = -1/4 and -1

Since x = tanθ

-1 =tanθ

θ = arctan(-1)

θ = -45degrees

Since tan is negative in the 2nd and fourth quadrant;

In the second quadrant

θ = 180 - 45 = 135degrees

In the fourth quadrant;

θ = 360 - 45 = 315degrees

Hence the required angles are 135 and 315degrees

3 0

3 years ago