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3 years ago

(2, 4) and (6, 12) find slope

Mathematics

1 answer:

Free_Kalibri [48]3 years ago

4 0

Answer:

Step-by-step explanation:

Use the formula: change of y over change of x.

When you do that, your systems should look like this:

x1 = 2

y1 = 4

x2 = 6

y2 = 12

After that you plug these number into the corresponding variable.

(12 - 4) / (6 - 2) = 8/4 which equals 2.

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A square garden plot measures 180 square feet. A second square garden plot measures 320 square feet. How many more feet of fence

REY [17]

The second garden plot will require 8√5 feet more fence than the first garden plot.

Further explanation:

In order to find the fence, we have to find the perimeter of both squares

So,

Area of Square 1: A1=180 square feet

Area of Square 2: A2=320 Square feet

Let x be the side of square 1:

Then,

$A_1=x^2\\180=x^2\\Taking\ square\ root\ on\ both\ sides\\\sqrt{180}=\sqrt{x^2}\\x=\sqrt{2*2*3*3*5}\\x=\sqrt{2^2*3^2*5}\\x=2*3\sqrt{5}\\x=6\sqrt{5}$

For second square:

Let y be the side of second square

$A_2=y^2\\320=y^2\\Taking\ square\ root\ on\ both\ sides\\\sqrt{320}=\sqrt{y^2}\\y=\sqrt{2*2*2*2*2*2*5}\\y=\sqrt{2^2*2^2*2^2*5}\\y=2*2*2\sqrt{5}\\y=8\sqrt{5}$

Perimeter of First Square:

$P_1=4x\\=4(6\sqrt{5})\\=24\sqrt{5}\ feet$

Perimeter of Second Square:

$P_2=4y\\=4(8\sqrt{5})\\=32\sqrt{5}\ feet$

The smaller perimeter will be subtracted from larger perimeter to find that how much more fence will be needed.

$P_2-P_1=32\sqrt{5}-24\sqrt{5}\\=(32-24)\sqrt{5}\\=8\sqrt{5}\ feet$

The second garden plot will require 8√5 feet more fence than the first garden plot.

Keywords: Radicals, Operations on Radicals

Learn more about radicals at:

#LearnwithBrainly

6 0

3 years ago

What is the area of this??? i put it at 14 points!

Basile [38]

Answer:

99in squared

Step-by-step explanation:

Bottom square is 3 by 3, middle is 9 by 6, and trapezoid is 9 by 4.

3*3+9*6+9*4 is 99

Hope this helps plz mark brainliest if correct :D

3 0

2 years ago

Find sin(u/2) if tan(u)=-7.721 and u is in quadrant 4

morpeh [17]

Answer:

.6601

Step-by-step explanation:

to find u, which we assume is an angle of course, we can take arctan of both sides. so arctan(-7.721) = -1.4420. this angle is in quadrant 4, but it is not in the range. In this instance you just go around the circle again, or add 2pi, which gets us 4.8412. You can check and you get the same answer if you take the tangent of either.

So now we have a number in the appropriate range in the appropriate quadrant. That means we have the correct angle. The question asks what is sine of that angle cut in half. so sin(4.8412/2), which gets you your answer. let em know if something didn't make sense.

7 0

3 years ago

In the rectangular prism shown below, which lines are parallel?

olga_2 [115]

Answer:

JN and LP...............

8 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E%7B%20%20%5Cpi%20%7D%20%5Ccos%28%20%5Ccot%28x%29%20%20%20%20-%20%2

Nikolay [14]

Replace x with π/2 - x to get the equivalent integral

$\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

but the integrand is even, so this is really just

$\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

Substitute x = 1/2 arccot(u/2), which transforms the integral to

$\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du$

There are lots of ways to compute this. What I did was to consider the complex contour integral

$\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz$

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

$\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}$

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

$\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}$

and it follows that

$\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}$

7 0

2 years ago