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Harlamova29_29 [7]
3 years ago
15

If the mass of a material is 45 grams and the volume of the material is 11 cm^3, what would the density of the material be?

Mathematics
1 answer:
Nuetrik [128]3 years ago
7 0

Answer:

Density of material would be 4.09 g/cm^3

units is g/cm^3

Step-by-step explanation:

Given: The mass of a material is 45 grams and the volume of the material is 11 cubic centimeter

Density is defined as mass per unit volume.

It is given by:

p= \frac{m}{V} where p is the density , m is the mass and V is the volume of the material respectively.

Here, Density is expressed in grams per centimeter cubed (g/cubic cm)

Here, m = 45 g , V = 11 cubic cm

We get;

p= \frac{45}{11} = 4.09 g/cm^3

therefore, density of a material would be, 4.09 g/cm^3

and its units is g/cm^3

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Step-by-step explanation:

Set up equation:

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Identify the interval on which the quadratic function is positive.
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Answer:

\textsf{1. \quad Solution:  $1 < x < 4$,\quad  Interval notation:  $(1, 4)$}

\textsf{2. \quad Solution:  $-2 < x < 4$,\quad  Interval notation:  $(-2, 4)$}

Step-by-step explanation:

<h3><u>Question 1</u></h3>

The intervals on which a <u>quadratic function</u> is positive are those intervals where the function is above the x-axis, i.e. where y > 0.

The zeros of the <u>quadratic function</u> are the points at which the parabola crosses the x-axis.  

As the given <u>quadratic function</u> has a negative leading coefficient, the parabola opens downwards.   Therefore, the interval on which y > 0 is between the zeros.

To find the zeros of the given <u>quadratic function</u>, substitute y = 0 and factor:

\begin{aligned}y&= 0\\\implies -7x^2+35x-28& = 0\\-7(x^2-5x+4)& = 0\\x^2-5x+4& = 0\\x^2-x-4x+4& = 0\\x(x-1)-4(x-1)&= 0\\(x-1)(x-4)& = 0\end{aligned}

Apply the <u>zero-product property</u>:

\implies x-1=0 \implies x=1

\implies x-4=0 \implies x=4

Therefore, the interval on which the function is positive is:

  • Solution:  1 < x < 4
  • Interval notation:  (1, 4)

<h3><u>Question 2</u></h3>

The intervals on which a <u>quadratic function</u> is negative are those intervals where the function is below the x-axis, i.e. where y < 0.

The zeros of the <u>quadratic function</u> are the points at which the parabola crosses the x-axis.  

As the given <u>quadratic function</u> has a positive leading coefficient, the parabola opens upwards.   Therefore, the interval on which y < 0 is between the zeros.

To find the zeros of the given <u>quadratic function</u>, substitute y = 0 and factor:

\begin{aligned}y&= 0\\\implies 2x^2-4x-16& = 0\\2(x^2-2x-8)& = 0\\x^2-2x-8& = 0\\x^2-4x+2-8& = 0\\x(x-4)+2(x-4)&= 0\\(x+2)(x-4)& = 0\end{aligned}

Apply the <u>zero-product property</u>:

\implies x+2=0 \implies x=-2

\implies x-4=0 \implies x=4

Therefore, the interval on which the function is negative is:

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Answer:

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Step-by-step explanation:

Use Pythagorean Theorem to solve this problem

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Answer:

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Step-by-step explanation:

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