All categories

3 years ago

Im struggling a lot with these problems can anyone help and show the WORK please?

Mathematics

1 answer:

Kay [80]3 years ago

5 0

Answer:

Step-by-step explanation:

I'll give a couple answers so you get the gist of things.

All you have to do is substitute 1/2 for x and 3/4 for y.

Ex - 30. x × y = 1/2 × 3/4 = 3/8

You might be interested in

Sally works 40 hours a week and makes $20 an hour, but her kids are in child care each day and the day care center charges her $

Maslowich

Answer:

$11.25

Step-by-step explanation:

Multiply the number of hours she works by how much she gets paid per hour.

800

Multiply 50 times 7 (50 being the amount charged for her kid and 7 being the days of the week)

350

Subtract the 350 from the 800

450

Divide the 450 by the number of hours worked per week

450/40

Answer: 11.25

So Sally actually makes 11.25 per hour.

5 0

3 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

2 years ago

I would appreciate it if someone would show me how to solve these y-intercept tables. I’ve been confused on them for a while, an

harina [27]

Answer:

Substitute the numbers below x by x in the equation given in the middle of the table and that will be the answer for y.

Step-by-step explanation:

For example, x= -1

y= x + 6

y= -1 + 6

y= 5

8 0

3 years ago

Pleaase help me with this question

damaskus [11]

Answer:A

Step-by-step explanation:

6 0

3 years ago

Tao wants to check that the expression 3(4x – 6) simplifies to 12x – 18. What is the value when 2 is substituted for x into both

timurjin [86]

Answer:

Step-by-step explanation:

3( 4x - 6)

Let x =2

3( 4*2 -6)

3(8-6)

3*2

Check

12x-18

12*2 -18

24-18

Both equal 6

4 0

3 years ago