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2 years ago

10

Choose the most appropriate translation of the English sentence.

1 answer:

11111nata11111 [884]2 years ago

4 0

Answer:

$25 per hour and you worked 12 hours=300

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Please help ASAP! 40 PTS

tester [92]

Answer:

see below

Step-by-step explanation:

1. < 4 is the exterior angle of a triangle given

2. <1 + <2 + <3 =180 basic property of triangles

3. <3 + <4 are a linear pair <3 and <4 are adjacent and form a straight line

4. <3 and <4 are supplementary angles supplementary angles are linear pairs

5. <3 + <4 = 180 The definition of supplementary angles are that they add to 180 degrees

6. <1 + <2 + <3 = <3+ <4 Substitute <3 + <4 for 180 in equation from step 2

7 <1 + <2 = <4 Subtract <3 from each side using the subtraction property of equality

4 0

3 years ago

Read 2 more answers

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

2 years ago

What is the image of (-9,-1) after a dilation by a scale of 2 centered at the origin

Dafna1 [17]

The answer is (-18, -2). Just multiply by 2!

8 0

2 years ago

What is the volume of the figure

Svetradugi [14.3K]

The volume of this figure is 660 cube centimetre

7 0

2 years ago

What is the solution to 3x -2y =1 and 4y = 7 + 3x

Delvig [45]

Given.

$3x-2y=1$ (1)

$4y=7+3x$ (2)

Re-arrange equation (2)

$(4y=7+3x)-4y$

$0=-4y+3x+7$

$(0=-4y+3x+7)-7$

$-7=-4y+3x$

Multiply equation (2) by -1

$-1(3x-4y=-7)$

$-3x+4y=7$ (3)

Combine like terms and solve.

$3x-3x-2y+4y=1+7$

$2y=8$

Divide by (2y) and solve for (y).

$\frac{2y}{2}=\frac{8}{2}$

$y=4$

Substitute (y) for equation (1).

$3x-2y=1$

$3x-2(4)=1$

$3x-8=1$

Add (8) to both sides.

$(3x-8)+8=(1)+8$

$3x=9$

Divide by (3).

$\frac{3x}{3}=\frac{9}{3}$

$x=3$

Check your work.

$3x-2y=1$

$3(3)-2(4)=1$

$9-8=1$

$1=1$

$4y=7+3x$

$4(4)=7+3(3)$

$16=7+9$

$16=16$

Your answers is $x=3$ and $y=4$

8 0

3 years ago