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Alexxx [7]
2 years ago
11

A rectangle has sides measuring (4x 5) units and (3x 10) units. part a: what is the expression that represents the area of the r

ectangle? show your work. (4 points) part b: what are the degree and classification of the expression obtained in part a? (3 points) part c: how does part a demonstrate the closure property for polynomials? (3 points)
Mathematics
1 answer:
Harrizon [31]2 years ago
5 0
  • a: The expression representing the area of the rectangle is given as,

(4x + 5)(3x + 10) or 12x² + 55x +50

  • b: The degree of the computed expression is 2 and it is classified as a quadratic expression.
  • c: The polynomial expression written in part a is closed under addition multiplication.

Forming the Expression for the Area of Rectangle

The formula for the area of a rectangle is given as,

Area, A = length × breadth

Here, the dimensions of the rectangle are given as,

length = (4x + 5)

breadth = (3x + 10)

∴ The equation for the area of the rectangle is,

A = (4x + 5)(3x + 10)

A = 12x² + 55x +50

Hence, 12x² + 55x +50 is the required expression.

Degree and Classification of the Expression

The highest power of a variable that is present in an expression is known as its degree.

Since the degree of the expression formed in part a is 2, it is classified to be a quadratic expression.

Closure Property For Polynomials

When the output is the same kind of object as the inputs, an expression is said to be closed. The expression obtained for the area of the rectangle is a combination of constants and variables closed under multiplication and addition.

Hence the expression formed in part a indicates at the closure property for polynomials.

Learn more about an expression here:

brainly.com/question/14083225

#SPJ4

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The exact value is found by making use of order of operations. The

functions can be resolved using the characteristics of quadratic functions.

Correct responses:

  • \displaystyle 1\frac{4}{7} \div \frac{2}{3} - 1\frac{5}{7} =\frac{9}{14}
  • x = -4, y = 12
  • When P = 1, V = 6
  • \displaystyle x = -3 \ or \ x = \frac{1}{2}

\displaystyle i. \hspace{0.1 cm} \underline{ f(x) = 2 \cdot \left(x - 1.25 \right)^2 + 4.875 }

ii. The function has a minimum point

iii. The value of <em>x</em> at the minimum point, is <u>1.25</u>

iv. The equation of the axis of symmetry is <u>x = 1.25</u>

<h3>Methods by which the above responses are found</h3>

First part:

The given expression, \displaystyle \mathbf{ 1\frac{4}{7} \div \frac{2}{3} -1\frac{5}{7}}, can be simplified using the algorithm for arithmetic operations as follows;

  • \displaystyle 1\frac{4}{7} \div \frac{2}{3} - 1\frac{5}{7} = \frac{11}{7}  \div \frac{2}{3} - \frac{12}{7} = \frac{11}{7} \times \frac{3}{2} - \frac{12}{7} = \frac{33 - 24}{14} =\underline{\frac{9}{14}}

Second part:

y = 8 - x

2·x² + x·y = -16

Therefore;

2·x² + x·(8 - x) = -16

2·x² + 8·x - x² + 16 = 0

x² + 8·x + 16 = 0

(x + 4)·(x + 4) = 0

  • <u>x = -4</u>

y = 8 - (-4) = 12

  • <u>y = 12</u>

Third part:

(i) P varies inversely as the square of <em>V</em>

Therefore;

\displaystyle P \propto \mathbf{\frac{1}{V^2}}

\displaystyle P = \frac{K}{V^2}

V = 3, when P = 4

Therefore;

\displaystyle 4 = \frac{K}{3^2}

K = 3² × 4 = 36

\displaystyle V = \sqrt{\frac{K}{P}

When P = 1, we have;

\displaystyle V =\sqrt{ \frac{36}{1} } = 6

  • When P = 1, V =<u> 6</u>

Fourth Part:

Required:

Solving for <em>x</em> in the equation; 2·x² + 5·x  - 3 = 0

Solution:

The equation can be simplified by rewriting the equation as follows;

2·x² + 5·x - 3 = 2·x² + 6·x - x - 3 = 0

2·x·(x + 3) - (x + 3) = 0

(x + 3)·(2·x - 1) = 0

  • \displaystyle \underline{x = -3 \ or\ x = \frac{1}{2}}

Fifth part:

The given function is; f(x) = 2·x² - 5·x + 8

i. Required; To write the function in the form a·(x + b)² + c

The vertex form of a quadratic equation is f(x) = a·(x - h)² + k, which is similar to the required form

Where;

(h, k) = The coordinate of the vertex

Therefore, the coordinates of the vertex of the quadratic equation is (b, c)

The x-coordinate of the vertex of a quadratic equation f(x) = a·x² + b·x + c,  is given as follows;

\displaystyle h = \mathbf{ \frac{-b}{2 \cdot a}}

Therefore, for the given equation, we have;

\displaystyle h = \frac{-(-5)}{2 \times 2} = \mathbf{ \frac{5}{4}} = 1.25

Therefore, at the vertex, we have;

k = \displaystyle f\left(1.25\right) = 2 \times \left(1.25\right)^2 - 5 \times 1.25  + 8 = \frac{39}{8} = 4.875

a = The leading coefficient = 2

b = -h

c = k

Which gives;

\displaystyle f(x) \ in \ the \ form \  a \cdot (x + b)^2 + c \ is \ f(x) = 2 \cdot \left(x + \left(-1.25 \right) \right)^2 +4.875

Therefore;

  • \displaystyle \underline{ f(x) = 2 \cdot \left(x -1.25\right)^2 + 4.875}

ii. The coefficient of the quadratic function is <em>2</em> which is positive, therefore;

  • <u>The function has a minimum point</u>.

iii. The value of <em>x</em> for which the minimum value occurs is -b = h which is therefore;

  • The x-coordinate of the vertex = h = -b =<u> 1.25 </u>

iv. The axis of symmetry is the vertical line that passes through the vertex.

Therefore;

  • The axis of symmetry is the line <u>x = 1.25</u>.

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