Given point (-6, -2) and a slope of 5, write an equation in slope-intercept form.

vladimir1956 [14]

13a) y = 0.45x

13b) 52 * 0.45 = 23.4

13c) I will let u figure this one out

7 0

3 years ago

A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. Find the dimensions of a norman

Yanka [14]

Answer:

$W\approx 8.72$ and $L\approx 15.57$ .

Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

$2L+W+\frac{1}{2}(2\pi r)=38$

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

$2L+W+\frac{1}{2}(2r\pi)=38$

$2L+W+\frac{\pi}{2}W=38$

Let us find area of window equation as:

$\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)$

$\text{Area}=W\cdot L+\frac{1}{2}(\pi (\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W^2}{4})$

$\text{Area}=W\cdot L+\frac{\pi}{8}W^2$

Now, we will solve for L is terms W from perimeter equation as:

$L=38-(W+\frac{\pi }{2}W)$

Substitute this value in area equation:

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

Since we need the area of window to maximize, so we need to optimize area equation.

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

$A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2$

Let us find derivative of area equation as:

$A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W$

$A'=38-2W-\pi W+\frac{\pi}{4}W$

$A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W$

$A'=38-2W-\frac{3\pi W}{4}$

To find maxima, we will equate first derivative equal to 0 as:

$38-2W-\frac{3\pi W}{4}=0$

$-2W-\frac{3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}*4=-38*4$

$-8W-3\pi W=-152$

$8W+3\pi W=152$

$W(8+3\pi)=152$

$W=\frac{152}{8+3\pi}$

$W=8.723210$

$W\approx 8.72$

Upon substituting $W=8.723210$ in equation $L=38-(W+\frac{\pi }{2}W)$ , we will get:

$L=38-(8.723210+\frac{\pi }{2}8.723210)$

$L=38-(8.723210+\frac{8.723210\pi }{2})$

$L=38-(8.723210+\frac{27.40477245}{2})$

$L=38-(8.723210+13.70238622)$

$L=38-(22.42559622)$

$L=15.57440378$

$L\approx 15.57$

Therefore, the dimensions of the window that will maximize the area would be $W\approx 8.72$ and $L\approx 15.57$ .

8 0

3 years ago

What is the slope of the line passing through the points (1, 2) and (5, 4)?

bulgar [2K]

<h2>SOLVING</h2>

$\Large\maltese\underline{\textsf{A. What is Asked}}$

What is the slope of the line passing through the point (1,2) and (5,4)

$\Large\maltese\underline{\textsf{This problem has been solved!}}$

Formula used, here $\bf{\dfrac{y2-y1}{x2-x1}$

_______________________________________________________

$\bf{\dfrac{4-2}{5-1}$ | simplify

$\bf{\dfrac{2}{4}$ | reduce

$\bf{\dfrac{1}{2}$

$\rule{300}{1.7}$

$\bf{Result:}$

$\bf{=Slope:\dfrac{1}{2}$

$\boxed{\bf{aesthetic\not101}}$

3 0

2 years ago

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What is the answer to this equation: 7.4 - 9 - 2r =18

lorasvet [3.4K]

We can get the answer to this equation by solving for the variable (r).

7.4 - 9 - 2r = 18

We want to get r on one side, so we want to push everything else to the other side.

-2r = 18 - 7.4 + 9

-2r = 19.6

2r = -19.6

r = -9.8

5 0

3 years ago

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This problem really confuses me, help?

OLga [1]

So k in the first problem would =1/2
and i guess in the second problem just put k in place of the first space and in the second space put 8 for x. y would then =4. Idk if I did this right, though, I'm honestly a little confused too.

3 0

3 years ago

Read 2 more answers