Answer:
happy and sad
Step-by-step explanation:
Answer:
Step-by-step explanation:
min: 8
max: 28
median: 22
1st Q: 12
3rd Q:26
Answer:
The small balloon bouquet uses 7 balloons and the large one uses
18 balloons.
Step-by-step explanation:
Let's say that small balloon bouquets are S and large balloon bouquets are L. For the graduation party the employee assembled 6 small bouquets and 6 large bouquets, the total number of balloon used is 150. To put the sentence into an equation will be:
6S + 6L= 150
S+L= 25 ----> 1st equation
For Father's Day, the employee uses 6 small bouquet and 1 large bouquet, the total number of balloons used is 60. The equation will be:
6S + 1L= 60
1L= 60- 6S ----> 2nd equation
We can solve the number of small balloon bouquet by substitute the 2nd equation into 1st. The calculation will be:
S+L = 25
S+ (60-6S)= 25
-5S= 25-60
-5S= -35
S= -35/-5
S=7
Then we can find L by substitute S value to 1st or 2nd equation.
S+L=25
7+L=25
L=18
Hope this helps ;)
Answer:
![\large\boxed{\ln\sqrt[3]{e^4}=\dfrac{4}{3}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B%5Cln%5Csqrt%5B3%5D%7Be%5E4%7D%3D%5Cdfrac%7B4%7D%7B3%7D%7D)
Step-by-step explanation:
![\text{Use}\\\\\sqrt[n]{a^m}=a^\frac{m}{n}\\\\\ln a^n=n\ln a\\\\\ln e=1\\-----------\\\\\ln\sqrt[3]{e^4}=\ln e^\frac{4}{3}=\dfrac{4}{3}\ln e=\dfrac{4}{3}\cdot1=\dfrac{4}{3}](https://tex.z-dn.net/?f=%5Ctext%7BUse%7D%5C%5C%5C%5C%5Csqrt%5Bn%5D%7Ba%5Em%7D%3Da%5E%5Cfrac%7Bm%7D%7Bn%7D%5C%5C%5C%5C%5Cln%20a%5En%3Dn%5Cln%20a%5C%5C%5C%5C%5Cln%20e%3D1%5C%5C-----------%5C%5C%5C%5C%5Cln%5Csqrt%5B3%5D%7Be%5E4%7D%3D%5Cln%20e%5E%5Cfrac%7B4%7D%7B3%7D%3D%5Cdfrac%7B4%7D%7B3%7D%5Cln%20e%3D%5Cdfrac%7B4%7D%7B3%7D%5Ccdot1%3D%5Cdfrac%7B4%7D%7B3%7D)
First, tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>), so if cos(<em>θ</em>) = 3/5 > 0 and tan(<em>θ</em>) < 0, then it follows that sin(<em>θ</em>) < 0.
Recall the Pythagorean identity:
sin²(<em>θ</em>) + cos²(<em>θ</em>) = 1
Then
sin(<em>θ</em>) = -√(1 - cos²(<em>θ</em>)) = -4/5
and so
tan(<em>θ</em>) = (-4/5) / (3/5) = -4/3
The remaining trig ratios are just reciprocals of the ones found already:
sec(<em>θ</em>) = 1/cos(<em>θ</em>) = 5/3
csc(<em>θ</em>) = 1/sin(<em>θ</em>) = -5/4
cot(<em>θ</em>) = 1/tan(<em>θ</em>) = -3/4
