All categories

3 years ago

Need help on all these questions!!

Mathematics

1 answer:

Yuki888 [10]3 years ago

5 0

Okay, so YZ = 3 cm. You have XM correct. And YM = 0.5.

Now, you have the midpoint M at the correct spot.

Use Pythagorean's theorem o find the length of AB. a² + b² = c² a=6, b=8.

6² = 36 8² = 64 36 + 64 = 100 AB = 10!

If AB = 10 then AM = 5 MB also = 5

If B is the midpoint of AC, C would be 12 rows down from A, and 16 columns to the right. The last spot where the line intersects.

There are your answers!

You might be interested in

(-8, -2) and (1, 9) find the slope

sergejj [24]

Answer:

11/9

Step-by-step explanation:

To find the slope

m = ( y2-y1)/(x2-x1)

= ( 9 - -2)/(1 - -8)

= ( 9+2)/ (1+8)

= 11/9

6 0

3 years ago

Help with the linear equations

IgorC [24]

Answer:

$y \: intercept = - 1 \frac{1}{2}$

Step-by-step explanation:

The y intercept form for the equation of a line is

$y = mx + c$

You should note that c represents the y-intercept of the line (where the line touches the y-axis)

$y = mx + c \\ \\ we \:were\: given \:the \:equation \:y = x - \frac{3}{2} \\ \\ therefore \: the\: value \:of \:c \:is\: - \frac{3}{2} \:or -1 \frac {1}{2}$

7 0

2 years ago

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 $x\approx3.36059492$ .

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

 $y\approx6-(1.68029746+2.63940507809)$

 $\boxed{y\approx1.68029746191}$

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

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

$\boxed{y\approx 1.68\ ft}$