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3 years ago

Is the relationship a function? (Show work if possible)

Mathematics

2 answers:

IgorLugansk [536]3 years ago

6 0

Step-by-step explanation:

This is not a function. This is not a function because the input (3) has 2 outputs which is 18 and 15.

I hope this helps! Have a good day!

MakcuM [25]3 years ago

4 0

Answer: No! This is not a function!

Step-by-step explanation:

Functions. Terms. Domain - The set of all inputs of a relation or function. Function - A relation in which each input has only one output. Often denoted f (x). Horizontal Line Test - If every horizontal line you can draw passes through only 1 point, x is a function of y.

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Which step is the most efficient way to find the solution of p + 5 > -13

jolli1 [7]

Answer:

subtracting 5 from both sides

Step-by-step explanation:

please mark brainliest lol :)

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3 years ago

Read 2 more answers

Massimo spent $90. riding the bus. Each bus ride cost $2.50. Write and solve an equation to determine how many times Massimo rod

Varvara68 [4.7K]

Answer:

$90 ÷ $2.50 = 36

Massimo rode the bus 36 times.

Please mark brainliest. Have a great day!

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3 years ago

A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six

kherson [118]

Answer:

x = 0.53 cm

Maximum volume = 1.75 cm³

Step-by-step explanation:

Refer to the attached diagram:

The volume of the box is given by

$V = Length \times Width \times Height \\\\$

Let x denote the length of the sides of the square as shown in the diagram.

The width of the shaded region is given by

$Width = 3 - 2x \\\\$

The length of the shaded region is given by

$Length = \frac{1}{2} (5 - 3x) \\\\$

So, the volume of the box becomes,

$V = \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V = \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V = \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V = \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\$

In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.

$\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15) \\\\18x^2 -38x + 15 = 0 \\\\$

We are left with a quadratic equation.

We may solve the quadratic equation using quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

$a = 18 \\\\b = -38 \\\\c = 15 \\\\$

$x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 + 19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\$