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4 years ago

In the right triangle shown, Angle A = 30° and AC = 18.

Mathematics

1 answer:

Komok [63]4 years ago

8 0

Answer:

Step-by-step explanation:

Angle A plus AC and that adds up too 48

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An isosceles right triangle has a hypotenuse of 16mm what would the area of the triangle be

olga2289 [7]

Answer: 64 square mm

Step-by-step explanation:

6 0

2 years ago

6³ ÷ 4 + 2 x 9(32 x 8 – 17 x 4).

inn [45]

ANSWER

3438

EXPLANATION:

PARENTHESIS FIRST

32 x 8 = 256

17 x 4 = 68

256 - 68 = 188

(now the equation is, 6³ ÷ 4 + 2 x 9 (188))

EXPONENTS NEXT

6³= 216

(now the equation is, 216 ÷ 4 + 2 x 9 (188))

MULTIPLICATION AND DIVISION NEXT (left to right)

216 ÷ 4 = 54

2 x 9 = 18

18 x 188 = 84

(now the equation is, 54 + 3384)

LASTLY ADDITION AND SUBTRACTION (left to right)

54 + 3384 = 3438

Ta-da!

8 0

3 years ago

a legal firm offers some services"pro bono" which means that they work for clients free of charge. the legal firm accepted 2% of

jeka94

252 divided by 2 percent = 12600 cases

4 0

4 years ago

Read 2 more answers

Carry out the following integrals, counterclockwise, around the indicated contour

Lady_Fox [76]

For the first integral, z = π/4 is a pole of order 3 and lies inside the contour |z| = 1. Compute the residue:

$\displaystyle \mathrm{Res}\left(\frac{e^z\cos(z)}{\left(z-\frac\pi4\right)^3}, z=\frac\pi4\right) = \lim_{z\to\frac\pi4}\frac1{(3-1)!} \frac{d^{3-1}}{dz^{3-1}}\left[e^z\cos(z)\right]$

We have

$\dfrac{d^2}{dz^2}[e^z\cos(z)] = -2e^z \sin(z)$

and so

$\displaystyle \mathrm{Res}\left(\frac{e^z\cos(z)}{\left(z-\frac\pi4\right)^3}, z=\frac\pi4\right) = - \lim_{z\to\frac\pi4} e^z \sin(z) = -\frac{e^{\pi/4}}{\sqrt2}$

Then by the residue theorem,

$\displaystyle \int_C \frac{e^z\cos(z)}{\left(z-\frac\pi4\right)^3} \, dz = 2\pi j \left(-\frac{e^{\pi/4}}{\sqrt2}\right) = \boxed{-\sqrt2\,\pi e^{\pi/4} j}$

For the second integral, z = 2j and z = j/2 are both poles of order 2. The second poles lies inside the rectangle, so just compute the residue there as usual:

$\displaystyle \mathrm{Res}\left(\frac{\cosh(2z)}{(z-2j)^2\left(z-\frac j2\right)^2}, z=\frac j2\right) = \lim_{z\to\frac j2}\frac1{(2-1)!} \frac{d^{2-1}}{dz^{2-1}}\left[\frac{\cosh(2z)}{(z-2j)^2}\right] = \frac{16\cos(1)-24\sin(1)}{27}j$

The other pole lies on the rectangle itself, and I'm not so sure how to handle it... You may be able to deform the contour and consider a principal value integral around the pole at z = 2j. The details elude me at the moment, however.

6 0

3 years ago

1.5d + 3.25 = 1 + 2.25d

Fittoniya [83]

The answer is d = 3 I am pretty sure

5 0

3 years ago