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4 years ago

Find the domain and range from the function listed below.

Mathematics

1 answer:

Phantasy [73]4 years ago

4 0

Answer:

Domain; [-1, 3)

Range; (-5, 4]

Step-by-step explanation:

The domain of a function is defined as the set of x-values for which the given function is real and defined. In order to find the domain of the graphed function, we have to determine the lowest and the highest x-values for which the function is defined. From the graph, the least value of x is -1 while the greatest x value is 3 but the function is not defined at this point. Therefore the domain of the function is;

[-1, 3)

On the other hand, the range refers to the set of y-values for which the function is real and defined. The least y-value from the given graph is -5 while the greatest y -value is 4. Therefore, the range of the graphed function is;

(-5, 4]

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2 sin^2 (x) + 3 sin (x) + 1 = 0<br> help me find the solution

Fed [463]

Answer:

Step-by-step explanation:

\[2~ sin^2 x+3sin x+1=0\]

\[2sin^2x+2sin x+sin x+1=0\]

2sinx(sin x+1)+1(sin x+1)=0

(sin x+1)(2 sin x+1)=0

either sin x+1=0

sin x=-1=sin 3π/2=sin (2nπ+3π/2)

x=2nπ+3π/2,where n is an integer.

or 2sin x+1=0

sin x=-1/2=-sin π/6=sin (π+π/6),sin (2π-π/6)=sin (2nπ+7π/6),sin (2nπ+11π/6)

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7 0

3 years ago

Kesha threw her baton up in the air from the marching band platform during practice. The equation h(t) = −16t² + 54t + 40 gives

lapo4ka [179]

Answer:

a) 40 feet

b) 54 ft/min

c) 4 mins

Step-by-step explanation:

Solution:-

- Kesha models the height ( h ) of the baton from the ground level but thrown from a platform of height hi.

- The function h ( t ) is modeled to follow a quadratic - parabolic path mathematically expressed as:

h ( t ) = −16t² + 54t + 40

Which gives the height of the baton from ground at time t mins.

- The initial point is of the height of the platform which is at a height of ( hi ) from the ground level.

- So the initial condition is expressed by time = 0 mins, the height of the baton h ( t ) would be:

h ( 0 ) = hi = -16*(0)^2 + 54*0 + 40

h ( 0 ) = hi = 0 + 0 + 40 = 40 feet

Answer: The height of the platform hi is 40 feet.

- The speed ( v ) during the parabolic path of the baton also varies with time t.

- The function of speed ( v ) with respect to time ( t ) can be determined by taking the derivative of displacement of baton from ground with respect to time t mins.

v ( t ) = dh / dt

v ( t )= d ( −16t² + 54t + 40 ) / dt

v ( t )= -2*(16)*t + 54

v ( t )= -32t + 54

- The velocity with which Kesha threw the baton is represented by tim t = 0 mins.

Hence,

v ( 0 ) = vi = -32*( 0 ) + 54

v ( 0 ) = vi = 54 ft / min

Answer: Kesha threw te baton with an initial speed of vo = 54 ft/min

- The baton reaches is maximum height h_max and comes down when all the kinetic energy is converted to potential energy. The baton starts to come down and cross the platform height hi = 40 feet and hits the ground.

- The height of the ball at ground is zero. Hence,

h ( t ) = 0

0 = −16t² + 54t + 40

0 = -8t^2 + 27t + 20

- Use the quadratic formula to solve the quadratic equation:

$t = \frac{27+/-\sqrt{27^2 - 4*8*(-20)} }{2*8}\\\\t = \frac{27+/-\sqrt{1369} }{16}\\\\t = \frac{27+/-37 }{16}\\\\t = \frac{27 + 37}{16} \\\\t = 4$

Answer: The time taken for the baton to hit the ground is t = 4 mins

3 0

3 years ago

Can someone please help me

xenn [34]

It’s y = (x + 1) ^2 - 9

4 0

2 years ago

Read 2 more answers

An architect made a scale drawing of a building. The scale of the drawing is 3 inches equals 50 feet. The actual building has a

melamori03 [73]

The height in the scale drawing was 250 inches.

7 0

3 years ago

Consider a line whose slope is 6 and which passes through the point (8.–2).

Zolol [24]

Answer:

$y=6(x-8)-2\qquad\text{point-slope form}$

$y=6x-50\qquad\text{slope-intercept form}$

Step-by-step explanation:

The equation of a line can be written in several forms. Two of the most-used forms are the point-slope and the slope-intercept forms.

The point-slope form requires to have one point (xo, yo) through which the line passes and the slope m. The equation expressed in this form is:

$y=m(x-xo)+yo$

The slope-intercept form requires to have the slope m and the y-intercept b, or the y-coordinate of the point where the line crosses the y-axis. The equation is:

$y=mx+b$

The line considered in the question has a slope m=6 and passes through the point (8,-2). These data is enough to find the point-slope form of the line:

$\boxed{y=6(x-8)-2\qquad\text{point-slope form}}$

To find the slope-intercept form, we operate the above equation:

$y=6x-48-2$

$\boxed{y=6x-50\qquad\text{slope-intercept form}}$

3 0

3 years ago