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3 years ago

The graph of a function is (always sometimes never) a line

Mathematics

2 answers:

vagabundo [1.1K]3 years ago

5 0

The answer is sometimes because it can be linear (line) or quadratic (curve). Hope this helps!

vfiekz [6]3 years ago

3 0

The answer is sometimes.

If the graph is linear, it would be a line, for example:
y = 2x - 5

If the graph is quadratic, it would be a curve, for example:
y = x^2 - 5x + 6

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Vinvika [58]

Answer:

Step-by-step explanation:

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3 years ago

PART A: Maria rented a coat at $285 for 3 days. If she rents the same coat for 60 days, she has to pay a total rent of $510. Wri

Dmitriy789 [7]

<span>1) Name the variables

Number of days: x

rent: y

2) state the initial points

   x           y
days        $

3           285

60           510

3) assume linear relation:

=> (y - yo) / (x - xo) = (y1 - yo) / (x1 - xo)

=> (y - 285) / (x - 3) = (510 - 285) / (60 - 3)

=> (y - 285) / (x - 3) = 225 / 57 = 75 / 19

=> 19 (y - 285) = 75 (x - 3)

=> 19y - 19*285 = 75x - 75*3

=> 19y - 75x = 5415 - 225

=> 19y - 75x = 5190

=> standar form = -75x + 19y = 5190

PART B: Write the equation obtained in Part A using function notation.

-75x + 19y = 5190

=> 19y = 5190 + 75x

=> y = 5190/19 + (75/19)x

=> function notation = f(x) = (75/19)x + 5190 / 19

PART C: Describe the steps to graph the equation obtained above on the coordinate axes. Mention the labels on the axes and the intervals.

1) Coordinate axes:

x: number of days

y: rent

2) draw the two given points: (3,285) and (60, 510)

3) draw the line that joins those points from the interception of the y-axis until some points further (60, 510).
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3 years ago

The float committee made a scale model of their float before they started construction. The float committee used the scale 1 inc

igor_vitrenko [27]

Answer:

Step-by-step explanation:

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3 years ago

Given the position of the particle, what the position(s) of the particle when it’s at rest

choli [55]

The position function of a particle is given by:

$X\mleft(t\mright)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t$

The velocity function is the derivative of the position:

$\begin{gathered} V(t)=\frac{2}{3}(3t^2)-\frac{9}{2}(2t)-18 \\ \text{Simplifying:} \\ V(t)=2t^2-9t-18 \end{gathered}$

The particle will be at rest when the velocity is 0, thus we solve the equation:

$2t^2-9t-18=0$

The coefficients of this equation are: a = 2, b = -9, c = -18

Solve by using the formula:

$t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

Substituting:

$\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}$

We have two possible answers: