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2 years ago

Can someone help me with this??

Mathematics

1 answer:

lukranit [14]2 years ago

4 0

10 40 400
20 30 600
25 25 625
35 15 525
40 60 2400
(mb if it’s incorrect)

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The width of a rectangle is 3ft less than the length. The perimeter is 62ft. Find the length and width of the rectangle.

Assoli18 [71]

I believe the length of the rectangle is 17 ft and the width is 14

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4 years ago

Veronica says that the expression

ELEN [110]

The person on top is correct

6 0

3 years ago

PLEASE HELP!!!!!!

harkovskaia [24]

-3y + 7 + 3(y - 1) = 2(2y + 6)
-3y + 7 + 3y - 3 = 4y + 12
4y = 4 - 12 = -8
y = -8/4 = -2.

-(x - 5)(x + 4) = (x + 7)(x - 6)
-(x^2 + 4x - 5x - 20) = x^2 - 6x + 7x - 42
-x^2 + x + 20 = x^2 + x - 42
2x^2 = 62
x^2 = 62/2 = 31
$x=\pm\sqrt{31}$

7 0

3 years ago

Read 2 more answers

List Three measurements with different units that are equal to five meters

77julia77 [94]

0.005 kilometers
500 centimeters
5,000 millimeters

7 0

4 years ago

israel started to solve a radical equation in this way: square root of x plus 6 − 4 = x square root of x plus 6 − 4 4 = x 4 squa

agasfer [191]

Lets solve our radical equation $\sqrt{x+6}-4=x$ step by step.

Step 1 add 4 to both sides of the equation:

$\sqrt{x+6} -4+4=x +4$

$\sqrt{x} +6=x+4$

Step 2 square both sides of the equation:

$\sqrt{x+6} =x+4$

$\sqrt{x+6}^2 =(x+4)^2$

$x+6=(x+4)^2$

Step 3 expand the binomial in the right hand side:

$x+6=x^2+8x+16$

Step 4 simplify the expression:

$0=x^2+8x-x+16-6$

$x^2+7x+10=0$

Step 5 factor the expression:

$(x+2)(x+5)=0$

Step 6 solve for each factor:

$x+2=0$ or $x+5=0$

$x=-2$ or $x=-5$

Now we are going to check both solutions in the original equation to prove if they are valid:

For $x=-2$

$\sqrt{x+6}-4=x$

$\sqrt{-2+6}-4=-2$

$\sqrt{4}-4=-2$

$2-4=-2$

$-2=-2$

The solution $x=2$ is a valid solution of the rational equation $\sqrt{x+6}-4=x$ .

For $x=-5$

$\sqrt{x+6}-4=x$

$\sqrt{-5+6}-4=-5$

$\sqrt{1}-4=-5$

$1-4=-5$

$-3\neq -5$

Since -3 is not equal to -5, the solution $x=-5$ is not a valid solution of the rational equation $\sqrt{x+6}-4=x$ ; therefore, $x=-5$ is an extraneous solution of the equation.

We can conclude that even all the algebraic procedures of Israel are correct, he did not check for extraneous solutions.

An extraneous solution of an equation is the solution that emerges from the algebraic process of solving the equation but is not a valid solution of the equation. Is worth pointing out that extraneous solutions are particularly frequent in rational equation.

8 0

3 years ago