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4 years ago

the area of a concrete rectangular pathway is 350m square and its perimeter pathway is 90 m.what is the length of the pathway?

1 answer:

5 0

X, y - the lengths of the sides of the pathway

The area of a rectangle is the product of the lengths of its sides. The area is 350 m².
$xy=350$

The perimeter of a rectangle is two times the sum of the lengths of its sides. The perimeter is 90 m.
$2(x+y)=90 \ \ \ \ \ \ |\div 2 \\
x+y=45 \ \ \ \ \ \ \ \ \ \ |-x \\
y=45-x$

Substitute 45-x for y in the first equation:
$x(45-x)=350 \\
45x-x^2=350 \\
-x^2+45x-350=0 \\
-x^2+10x+35x-350=0 \\
-x(x-10)+35(x-10)=0 \\
(-x+35)(x-10)=0 \\
-x+35=0 \ \lor \ x-10=0 \\
x=35 \ \lor \ x=10$

$y=45-x \\
\hbox{for } x=35: \\
y=45-35=10 \\ \\
\hbox{for } x=10: \\
y=45-10=35$

The dimensions of the pathway are 10 m by 35 m.
The length of the pathway is 35 m.

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Data and Calculations:

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2 years ago

Show all work to solve the equation for x. If a solution is extraneous, be sure to identify it in your final answer.

IRISSAK [1]

Answer:

There is one extraneous solution at $x = 4$

The only solution is $x=7$

Step-by-step explanation:

Given:

The equation to solve is given as:

$\sqrt{x-3}+5=x$

In order to solve, we will first isolate the radical part.

<em>Adding -5 on both the sides, we get:</em>

$\sqrt{x-3}+5-5=x-5\\\sqrt{x-3}=x-5$

<em>Now, squaring both the sides, we get:</em>

$(\sqrt{x-3})^2=(x-5)^2\\\\x-3=x^2+25-10x\\\\x^2-10x-x+25+3=0\\\\x^2-11x+28=0\\\\(x-4)(x-7)=0\\\\x=4\ or\ x=7$

Now, let us check whether the calculated values satisfy the equations or not.

Plug in 4 and 7 for 'x' and check the value of the equation.

At $x=4$ ,

$\sqrt{4-3}+5=4\\\sqrt1+5=4\\1+5=4\\6=4$

But, 6 is not equal to 4 and hence 4 is not a solution to the given equation. It is an extraneous solution for the given equation.

Now, at $x=7$

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Hence, there is one extraneous solution at $x = 4$

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