Please help me pleaseeee I’ll mark brainliest for the first person who answers no joke

Given the points (3, 2) and (6, 4), which of the following are true about the line passing through these points?

Artist 52 [7]

I don't know what 'which of the following' is referring to, but the line has a positive slope, goes through the origin, and has a slope of 2/3.

5 0

3 years ago

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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>

3 0

3 years ago

C = 3/4y + b, solve for b

12345 [234]

Answer:

b = c - 3/4y

Step-by-step explanation:

move 3/4y to the other side with c:

c - 3/4y = b

b = c - 3/4y

4 0

2 years ago

Find constants a and b so that the minimum for the parabola f(x)=x^2+ax+b is at the point (6,7)

Blizzard [7]

To find the minimum, take the derivative of the function and equate to zero:

f(x) = y = x² + ax + b
dy/dx = 2x + a = 0
Substitute x =6:
a = -2x
a = -2(6)
a = -12

Then, substitute x = 6, y = 7 and a = -12 to find b.

7 = (6)² -12(6) + b
7 + 36 = b
b = 43

Thus, a = -12 and b = 43

4 0

3 years ago

Lorenzo drew 3 triangles, 4 rectangles, and 7 arrows in a set of 20 shapes. The only other type of shape in the set is a circle.

djyliett [7]

Answer:

$\large\boxed{6/20}$

Step-by-step explanation:

In this question, we're trying to find how many circles are there and represent it as a fraction.

We know that he drew:

3 triangles
4 rectangles
7 arrows
Set of 20 shapes

We need to find how many circles are there.

3 tiangles + 4 rectangles + 7 arrows = 14 shapes

We know that there is suppose to be 20 shapes, this means that the remaining amount would be circles (6 circles)

We now know that there are 6 circles and 20 shapes in total.

This means that the fraction that would represent circles is 6/20

<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>

8 0

3 years ago

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