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3 years ago

Which value must be added to the expression x^2+12x to make it a perfect square trinomial a:6 b:36 c:72 d:144

Mathematics

2 answers:

Ksivusya [100]3 years ago

7 0

The answer would be 36. Why?

Because in the equation we would need to do (x + 6)(x + 6) to get a perfect square trinomial.

And 6 * 6 is 36.

So the whole equation would be: x^2 + 12 + 36.

Hope this helped! c:

mariarad [96]3 years ago

4 0

The quadratic expression uses the following formula:

$ax^2 + bx + c$

The expression is missing the variable of c, and it is a perfect square trinomial.

To find c in a perfect square trinomial, we will use the following formula:

$c = (\frac{b}{2}})^2$

The value of b in this expression is 12. Plug this value into the formula:

$\frac{12}{2} = 6$

$6^2 = \boxed{36}$

36 must be added to make this trinomial a perfect square.

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A normal window is constructed by adjoining a semicircle to the top of an ordinary rectangular window, (see figure ) The perimet

Dima020 [189]

Let's find the perimeter of the window.

The bottom side is $x$ . The left and right sides make $2y$ .
The perimeter of a circle is $2\pi r$ , so the perimeter of a semicircle must be $\pi r$ , The radius is $\frac{1}2x$ , so that gives $\frac{1}2\pi x$ for the curve. All of that is equal to 12.

$x+2y+\frac{1}2\pi x=12$

We only want to use one variable to create the area formula, so let's solve for $y$ .

$2y=12-x-\frac{1}2\pi x$

$y=6-\frac{1}2x-\frac{1}4\pi x$

Now that we have a value for $y$ in terms of $x$ , we can find the area in terms of $x$ .

The area of the rectangle is going to be $xy$ , which then becomes

$A_r=x(6-\frac{1}2x-\frac{1}4\pi x$

$A_r=6x-\frac{1}2x^2-\frac{1}4\pi x^2$

The area of the semicircle is going to be $\frac{1}2\pi r^2$ .

Since $r=\frac{1}2x$ , $A_{sc}=\frac{1}2\pi (\frac{1}2x)^2$ .

$A_{sc}=\frac{1}2\pi \frac{1}4x^2$

$A_{sc}=\frac{1}8\pi x^2$

Now let's add the areas of the rectangle and semicircle.

$A=A_r+A_{sc}$

$A=6x-\frac{1}2x^2-\frac{1}4\pi x^2+\frac{1}8\pi x^2$

$A=6x-\frac{1}2x^2-\frac{1}8\pi x^2$

If you wanted to factor out $\frac{1}8$ like you did, this would become

$\boxed{A(x)=\frac{1}8(48x=4x^2-\pi x^2)}$

Now what we want to do is find what $x$ is when $A$ is at its highest point, Once we have the value for $x$ we can also find the value for $y$ , of course.

Let's put our equation in the general form of a quadratic.

$A(x)=(-\frac{1}2-\frac{1}8\pi )x^2+6x$

Now we can use the vertex formula $x=\frac{-b}{2a}$ .
( $a$ and $b$ refer to $ax^2+bx+c$ .)

$x=\frac{-6}{2(-\frac{1}2-\frac{1}8\pi)}$

$x=\frac{-6}{-\frac{1}4\pi -1}$

$x=\frac{-24}{-\pi -4}$

$\boxed{x=\frac{24}{\pi +4}}$

Now let's plug that in for $y=6-\frac{1}2x-\frac{1}4\pi x$ .

Since our final answers are in decimal form and not exact form, we can make our lives a little easier here and just use $x\approx3.36059492$ .

$y\approx6-\frac{1}2(3.36059492)-\frac{1}4\pi(3.36059492)$

<span> $y\approx6-(1.68029746+2.63940507809)$

</span><span> $\boxed{y\approx1.68029746191}$

Let's take our answers for $x$ and $y$ and round to 2 decimal places.

$\boxed{x\approx 3.36\ ft}$

$\boxed{y\approx 1.68\ ft}$ </span>

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Bumek [7]

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