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2 years ago

Please help :) I’m really in need of assistance and would appreciate it.

Mathematics

1 answer:

BaLLatris [955]2 years ago

4 0

Answer:

$D) 0.5x^3y + 4x^2 y^2 - 2xy^3$

Step-by-step explanation:

Set up an equation to model the situation,

$(-0.5xy)(x^2 + 8xy - 4y^2)$

Distribute, multiply each term in the parenthesis with the term outside of the parenthesis. Don't forget the laws of exponents, when one multiplies two exponents, with the same base, then one adds the two exponents. A number with no exponential term is two the first degree meaning that it is to the power of one.

$-0.5x^3 y - 4x^2 y^2 + 2xy^3$

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A rectangle has width that is 6 meters less than the length. The area of the rectangle is 280 square meters. Find the dimensions

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Dimensions are length 20 meter and width 14 meter

Solution:

Let "a" be the length of rectangle

Let "b" be the width of rectangle

Given that,

A rectangle has width that is 6 meters less than the length

Width = length - 6

b = a - 6

The area of the rectangle is 280 square meters

The area of the rectangle is given by formula:

$Area = length \times width$

Substituting the values we get,

$Area = a \times (a-6)\\\\280 = a^2-6a\\\\a^2-6a -280=0$

Solve the above equation by quadratic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-6,\:c=-280:\quad a_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:1\left(-280\right)}}{2\cdot \:1}$

$a =\frac{6 \pm \sqrt{36+1120}}{2}\\\\a = \frac{6 \pm \sqrt{1156}}{2}\\\\a = \frac{6 \pm 34}{2}\\\\Thus\ we\ have\ two\ solutions\\\\a = \frac{6+34}{2} \text{ or } a = \frac{6-34}{2}\\\\a = 20 \text{ or } a = -14$

Since, length cannot be negative, ignore a = -14

Thus solution of length is a = 20

Therefore,

width = length - 6

width = 20 - 6 = 14

Thus dimensions are length 20 meter and width 14 meter

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