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Lisa, an experience shipping clerk, can fill a certain order in 10 hours. Bill, a new clerk, needs 13 hours to do the same job.

Lelu [443]

I think 11 hours and 30 minutes

7 0

4 years ago

Find e^cos(2+3i) as a complex number expressed in Cartesian form.

ozzi

Answer:

The complex number $e^{\cos(2+31)} = \exp(\cos(2+3i))$ has Cartesian form

$\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2)$ .

Step-by-step explanation:

First, we need to recall the definition of $\cos z$ when $z$ is a complex number:

$\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}$ .

Then,

$\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}$ . (I)

Now, recall the definition of the complex exponential:

$e^{z}=e^{x+iy} = e^x(\cos y +i\sin y)$ .

So,

$e^{2i-3} = e^{-3}(\cos 2+i\sin 2)$

$e^{-2i+3} = e^{3}(\cos 2-i\sin 2)$ (we use that $\sin(-y)=-\sin y)$ .

Thus,

$e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)$

Now we group conveniently in the above expression:

$e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2$ .

Now, substituting this equality in (I) we get

$\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2$ .

Thus,

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)$

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right]$ .

5 0

3 years ago

A, B, and C are points of tangency. CQ = 5, PQ = 10, and PR = 14. What is the perimeter

igomit [66]

*see attachment for diagram

Answer:

Perimeter = 38

Step-by-step explanation:

Recall: when two tangents are drawn to meet at a point outside a circle, the segments of the two tangents are congruent.

Given,

CQ = 5

PQ = 10

PR = 14

Perimeter of ∆PQR = RC + CQ + QB + BP + PA + AR

CQ = QB = 5 (tangents drawn from an external point)

BP = PQ - QB

BP = 10 - 5 = 5

BP = PA = 5 (tangents drawn from an external point)

AR = PR - PA

AR = 14 - 5 = 9

AR = RC = 9 (tangents drawn from an external point)

✔️Perimeter of ∆PQR = RC + CQ + QB + BP + PA + AR

= 9 + 5 + 5 + 5 + 5 + 9

Perimeter = 38

7 0

3 years ago

Evaluate: [(21 + 6) − 32] ÷ 9.2 A) 4. B) 4.6. C) 1. D) 52.

Dima020 [189]

[(21 + 6) - 32] : 9.2 = (27 - 32) : 9.2 = -5 : 9.2 = -46

5 0

3 years ago

What is the value of x?<br> A.5<br> B.2.5<br> C.7.5<br> D.10

Pavlova-9 [17]

$\text{We have proportion:}\\\\\dfrac{x}{x-2}=\dfrac{x+5}{x+1}\qquad\text{cross multiply}\\\\x(x+1)=(x-2)(x+5)\qquad\text{use distributive property and FOIL method}\\\\(x)(x)+(x)(1)=(x)(x)+(x)(5)+(-2)(x)+(-2)(5)\\\\x^2+x=x^2+5x-2x-10\qquad\text{subtract}\ x^2\ \text{from both sides}\\\\x=3x-10\qquad\text{subtract 3x from both sides}\\\\-2x=-10\qquad\text{divide both sides by (-2)}\\\\x=5\\\\Answer:\ \boxed{A.\ 5}$

4 0

3 years ago