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Monica [59]
2 years ago
15

Pls help!! And show work thank you

Mathematics
2 answers:
leva [86]2 years ago
7 0

Answer:13 and 12

Step-by-step explanation: the pattern is add 2 then minus one. But then the 2 continues to 3 4 5 and so on. So u add 5 to 8 which gives 13 then minus 1 again to get 12

Nuetrik [128]2 years ago
6 0

Answer:

13,12

Step-by-step explanation:

from 2 it goes up by 2 to get 4 then subtracts by 1 to get 3, then goes up by 3 to get 6 then subtract one to get 5, then it goes up by 4 to get 9 then goes down by 1 to get 8.

Following the pattern, go up by 5 to get 13 then subtract 1 to get 12

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Answer:

0.77 ounces

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If one side of a square notebook measures 20 cm, what is the area of the front cover of the notebook?
jarptica [38.1K]

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400 cm squared

Step-by-step explanation:

Since a square has all equal sides, this would mean the the all the other sides are 20 as well. To get the area, you would just multiply 20 by 20 to get 400. :)

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Find the absolute value of the complex number 5 - 4i.
Damm [24]
I answered a similar question for someone else. The absolute value of a complex number is called the modulus. This represents the DISTANCE to the origin from the point on the imaginary + real plane. 
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The two equations will intersect at which of the following coordinates : y= -x + 3 y= 1/2x
kvv77 [185]
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3 years ago
xand y are light house. y being 20km due east of x and from a ship due south of x the bearing of y was 055°. what is the distanc
leonid [27]

Answer:

I) |xz| ≈ 28.6 km

II) |yz| ≈ 34.8 km

Step-by-step explanation:

Let's assume that the position of ship due south of x is z (aà pictor representation of the question is attached)

|xy| = 20 km, |xz| = ?, |yz| = ?, θ(y) = 55°

Using Trigonometric ratio - SOHCAHTOA

I) Tan θ = |xz| ÷ |xy| ⇒ Tan 55° = |xz| ÷ 20

|xz| = 20 * Tan 55 = 20 * 1.428

|xz| = 28.56 km

|xz| ≈ <u>28.6 km</u>

<u />

II) Cos θ = |xy| ÷ |yz| ⇒ Cos 55° = 20 ÷ |yz|

|yz| * Cos 55° = 20 ⇒ |yz| = 20 ÷ Cos 55°

|yz| = 20 ÷ 0.574 = 34.84 km

|yz| ≈ <u>34.8 km</u>

4 0
3 years ago
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