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Maru [420]
3 years ago
14

What’s the slope?? Please help!

Mathematics
2 answers:
frutty [35]3 years ago
7 0
(change in y divided by change in x)
9966 [12]3 years ago
4 0

Answer:

I believe the slope of the line is 5

Step-by-step explanation:

5 + 5 = 10

10 + 5 + 15

15 + 5 = 20

Hope this helps :)

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Find the greatest common factor and the least common multiple of 12 and 20 explain
klasskru [66]

Answer:

GCF: 4

LCM: 60

4 0
3 years ago
PLZ HELP MEEEEEE!!! I have a test and i need help plz help me.​
beks73 [17]

Answer:

1.) different, subtract, 14

2.) same, add, -33

3.) same, add, -70

4.) different, subtract, 0

5.) different, subtract, -10

6.) different, add, 12

7.) same, subtract, 3

8.) different, subtract, -4

9.) different, subtract, -4

10.) 14

11.) Mark has more, he has $14 more

12.) 9

13) -13 - 12 = -25

14) 200 - 40 = 60

15) 12 - 3 = 8

6 0
3 years ago
PLEASE HURRY ITS A TEST
AnnyKZ [126]
8+5+8=21 coins in total.
8/21*7/20=
56/420=
2/15. Hope this helps, and PM me if something's not clear!
4 0
3 years ago
PLEASE HELP<br> If p(-3) = 18, how would you represent this as an ordered pair?
disa [49]

An ordered pair is written as (x, y)

P(-3) = 18

The -3 is the x value and the 18 is the y value

Answer: (-3,18)

6 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
3 years ago
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