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alexandr1967 [171]
3 years ago
8

the length of a rectangle is 5 centimeters less than six times its width. its area is 14 square centimeters. find the dimensions

of the rectangle.
Mathematics
2 answers:
pychu [463]3 years ago
5 0

Answer:

The length is 2.2 centimetres   and the width is 6.5 centimetres

Step-by-step explanation:

Let l be the length of the rectangle and w be the width of the rectangle.

First we will write the statement mathematically

"the length of a rectangle is 5 centimetres less than six times its width" can be written as ;

l = 5 - 6w ---------------------- ----------------------------------------(1)

"its area is 14 square centimetres" can be written as;

l × w =  14 ---------------------------------------------------------------(2)

substitute equation (1) in equation (2)

(5-6w)w = 14

5w - 6w² = 14

The equation can be re-arrange as;

6w² - 5w -14=0

Using formular method to solve;

] = 6 b = -5   c = -14

x =- b ±√b² - 4ac   /2

x = -6  ±√25 + 336

x = -6± √361

x = -6±19/2

Either x = -6 + 19 or  x = -6 -19

x=  13/2   or   x = -25/2

There is no such thing as negative length, so we will take w = 13/2

w = 13/2

Substitute w = 13/2 in equation (2)

l × w = 14

l×13/2 = 14

13l / 2 = 14

multiply both-side by 2

13l = 28

Divide both-side by 13

l ≈ 2.2

Therefore,

The width = 13/2 = 6.5  and the length is 2.2

V125BC [204]3 years ago
3 0
Not sure if you are familiar with the quadratic equation, but see picture
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Rationalize the denominator of $\frac{5}{2+\sqrt{6}}$. The answer can be written as $\frac{A\sqrt{B}+C}{D}$, where $A$, $B$, $C$
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Answer:

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Step-by-step explanation:

Given:

$\frac{5}{2+\sqrt{6}}$

To find:

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$\frac{A\sqrt{B}+C}{D}$

<u>Solution:</u>

We can see that the resulting expression does not contain anything under \sqrt (square root) so we need to rationalize the denominator to remove the square root from denominator.

The rule to rationalize is:

Any term having square root term in the denominator, multiply and divide with the expression by changing the sign of square root term of the denominator.

Applying this rule to rationalize the given expression:

\dfrac{5}{2+\sqrt{6}} \times \dfrac{2-\sqrt6}{2-\sqrt6}\\\Rightarrow \dfrac{5 \times (2-\sqrt6)}{(2+\sqrt{6}) \times (2-\sqrt6)} \\\Rightarrow \dfrac{10-5\sqrt6}{2^2-(\sqrt6)^2}\ \ \ \ \   (\because \bold{(a+b)(a-b)=a^2-b^2})\\\Rightarrow \dfrac{10-5\sqrt6}{4-6}\\\Rightarrow \dfrac{10-5\sqrt6}{-2}\\\Rightarrow \dfrac{-5\sqrt6+10}{-2}\\\Rightarrow \dfrac{5\sqrt6-10}{2}

Comparing the above expression with:

$\frac{A\sqrt{B}+C}{D}$

A = 5, B = 6 (Not divisible by square of any prime)

C = -10

D = 2 (positive)

GCD of A, C and D is 1.

So, A +B+C+D = 5+6-10+2 = \bold3

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Step-by-step explanation:

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