They are the same angle and on opposite sides of the parallelogram
Answer: -√1/✓41
Step-by-step explanation:
Let y be the irrational number that will be multiplied by -√41 to get the product that equals 1.
y × (-√41) = 1
We then solve for y, by dividing through with -√41. This will be:
[y × (-√41)]/-√41 = 1/-√41
y = -1/√41
y = -√1/✓41
The irrational number is negative root one over root forty one.
The first one is 24, and the second one is 12
Answer:
Step-by-step explanation:
The slope intercept form of an equation of a line is y = mx + b, where m is the slope and b is the y-intercept (the value of y when x=0).
Since the slope is (1/2), we can write:
y = (1/2)x + b
We want a value of b such that it forces the line to go through point (-10,9). Enter that point in the equation and solve for b:
y = (1/2)x + b
9 = (1/2)*(-10) + b
9 = -5 + b
b = 14
The eqyuation of a line with a slope of (1/2) and goes through point (-10,9) is:
y = (1/2)x + 14
See attachment.
Answer:
![\sin L = 0.60](https://tex.z-dn.net/?f=%5Csin%20L%20%3D%200.60)
![tan\ N = 1.33](https://tex.z-dn.net/?f=tan%5C%20N%20%3D%201.33)
![\cos L = 0.80](https://tex.z-dn.net/?f=%5Ccos%20L%20%3D%200.80)
![\sin N = 0.80](https://tex.z-dn.net/?f=%5Csin%20N%20%3D%200.80)
Step-by-step explanation:
Given
See attachment
From the attachment, we have:
![MN = 6](https://tex.z-dn.net/?f=MN%20%3D%206)
![LN = 10](https://tex.z-dn.net/?f=LN%20%3D%2010)
First, we need to calculate length LM,
Using Pythagoras theorem:
![LN^2 = MN^2 + LM^2](https://tex.z-dn.net/?f=LN%5E2%20%3D%20MN%5E2%20%2B%20LM%5E2)
![10^2 = 6^2 + LM^2](https://tex.z-dn.net/?f=10%5E2%20%3D%206%5E2%20%2B%20LM%5E2)
![100 = 36 + LM^2](https://tex.z-dn.net/?f=100%20%3D%2036%20%2B%20LM%5E2)
Collect Like Terms
![LM^2 = 100 - 36](https://tex.z-dn.net/?f=LM%5E2%20%3D%20100%20-%2036)
![LM^2 = 64](https://tex.z-dn.net/?f=LM%5E2%20%3D%2064)
![LM = 8](https://tex.z-dn.net/?f=LM%20%3D%208)
Solving (a): ![\sin L](https://tex.z-dn.net/?f=%5Csin%20L)
![\sin L = \frac{Opposite}{Hypotenuse}](https://tex.z-dn.net/?f=%5Csin%20L%20%3D%20%5Cfrac%7BOpposite%7D%7BHypotenuse%7D)
![\sin L = \frac{MN}{LN}](https://tex.z-dn.net/?f=%5Csin%20L%20%3D%20%5Cfrac%7BMN%7D%7BLN%7D)
Substitute values for MN and LN
![\sin L = \frac{6}{10}](https://tex.z-dn.net/?f=%5Csin%20L%20%3D%20%5Cfrac%7B6%7D%7B10%7D)
![\sin L = 0.60](https://tex.z-dn.net/?f=%5Csin%20L%20%3D%200.60)
Solving (b): ![tan\ N](https://tex.z-dn.net/?f=tan%5C%20N)
![tan\ N = \frac{Opposite}{Adjacent}](https://tex.z-dn.net/?f=tan%5C%20N%20%3D%20%5Cfrac%7BOpposite%7D%7BAdjacent%7D)
![tan\ N = \frac{LM}{MN}](https://tex.z-dn.net/?f=tan%5C%20N%20%3D%20%5Cfrac%7BLM%7D%7BMN%7D)
Substitute values for LM and MN
![tan\ N = \frac{8}{6}](https://tex.z-dn.net/?f=tan%5C%20N%20%3D%20%5Cfrac%7B8%7D%7B6%7D)
![tan\ N = 1.33](https://tex.z-dn.net/?f=tan%5C%20N%20%3D%201.33)
Solving (c): ![\cos L](https://tex.z-dn.net/?f=%5Ccos%20L)
![\cos L = \frac{Adjacent}{Hypotenuse}](https://tex.z-dn.net/?f=%5Ccos%20L%20%3D%20%5Cfrac%7BAdjacent%7D%7BHypotenuse%7D)
![\cos L = \frac{LM}{LN}](https://tex.z-dn.net/?f=%5Ccos%20L%20%3D%20%5Cfrac%7BLM%7D%7BLN%7D)
Substitute values for LN and LM
![\cos L = \frac{8}{10}](https://tex.z-dn.net/?f=%5Ccos%20L%20%3D%20%5Cfrac%7B8%7D%7B10%7D)
![\cos L = 0.80](https://tex.z-dn.net/?f=%5Ccos%20L%20%3D%200.80)
Solving (d): ![\sin N](https://tex.z-dn.net/?f=%5Csin%20N)
![\sin N = \frac{Opposite}{Hypotenuse}](https://tex.z-dn.net/?f=%5Csin%20N%20%3D%20%5Cfrac%7BOpposite%7D%7BHypotenuse%7D)
![\sin N = \frac{LM}{LN}](https://tex.z-dn.net/?f=%5Csin%20N%20%3D%20%5Cfrac%7BLM%7D%7BLN%7D)
Substitute values for LM and LN
![\sin N = \frac{8}{10}](https://tex.z-dn.net/?f=%5Csin%20N%20%3D%20%5Cfrac%7B8%7D%7B10%7D)
![\sin N = 0.80](https://tex.z-dn.net/?f=%5Csin%20N%20%3D%200.80)