Okay so what we know is that there are 6 sides.
—————-————————————————————
3separate groups have the same measurements. So really 3 groups of 2 that have the same measurements.
—————————————————————
2 sides are 3*2 —————>an area of 6
2 sides are 3*2—————>an area of 6
2 sides are 2*2—————>an area of 4
—————————————————————so, we add those #s up and get 6+6+4 which equals 16 as your answer.
Based on the amount that Nick is left with at the end of the month, the amount he earns is $2,400
<h3>What does Nick earn?</h3>
The amount of $300 that is left is 25% of 50% of Nick's income as the other 75% of the 50% went to paying for the car and other expenses.
25% of 50% is:
= 25% x 50%
= 12.5%
The amount Nick makes per month is:
= Amount remaining / Amount remaining in percentage terms
= 300 / 12.5%
= $2,400
Find out more on percentages at brainly.com/question/843074
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Step-by-step explanation:
Mark each mark by 15 mins add an extra tick at the time section
the corner before the first 15 min mark make a tilted line to 15 mins and 30,000 feet draw a straight line till 2 hours and 15 mins draw titled down to 2 hours and 30 mins
Answer:
![\tan(\theta)=\frac{-\sqrt{5}}{2}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%3D%5Cfrac%7B-%5Csqrt%7B5%7D%7D%7B2%7D)
Step-by-step explanation:
Since we are in quadrant 2, sine is positive. Since sine is positive and cosine is negative, then tangent is negative.
Now I'm going to find the sine value of this angle given using one of the Pythagorean Identities, namely
.
If given
, then we have
by substitution of
.
Let's solve:
for
.
![\sin^2(\theta)+\frac{4}{9}=1](https://tex.z-dn.net/?f=%5Csin%5E2%28%5Ctheta%29%2B%5Cfrac%7B4%7D%7B9%7D%3D1)
Subtract 4/9 on both sides:
![\sin^2(\theta)=1-\frac{4}{9}](https://tex.z-dn.net/?f=%5Csin%5E2%28%5Ctheta%29%3D1-%5Cfrac%7B4%7D%7B9%7D)
Simplify:
![\sin^2(\theta)=\frac{5}{9}](https://tex.z-dn.net/?f=%5Csin%5E2%28%5Ctheta%29%3D%5Cfrac%7B5%7D%7B9%7D)
Square root both sides:
![\sin(\theta)=\sqrt{\frac{5}{9}}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%29%3D%5Csqrt%7B%5Cfrac%7B5%7D%7B9%7D%7D)
![\sin(\theta)=\frac{\sqrt{5}}{\sqrt{9}}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%29%3D%5Cfrac%7B%5Csqrt%7B5%7D%7D%7B%5Csqrt%7B9%7D%7D)
![\sin(\theta)=\frac{\sqrt{5}}{3}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%29%3D%5Cfrac%7B%5Csqrt%7B5%7D%7D%7B3%7D)
===========
![\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{\frac{\sqrt{5}}{3}}{\frac{-2}{3}}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%3D%5Cfrac%7B%5Csin%28%5Ctheta%29%7D%7B%5Ccos%28%5Ctheta%29%7D%3D%5Cfrac%7B%5Cfrac%7B%5Csqrt%7B5%7D%7D%7B3%7D%7D%7B%5Cfrac%7B-2%7D%7B3%7D%7D)
Multiplying top and bottom by 3 gives:
![\tan(\theta)=\frac{\sqrt{5}}{-2}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%3D%5Cfrac%7B%5Csqrt%7B5%7D%7D%7B-2%7D)
I'm going to move the factor of -1 to the top:
![\tan(\theta)=\frac{-\sqrt{5}}{2}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%3D%5Cfrac%7B-%5Csqrt%7B5%7D%7D%7B2%7D)