Answer:
sorry for wrong answer
Step-by-step explanation:
Labor unions arose in the nineteenth century as increasing numbers of Americans took jobs in factories, mines, and mills in the growing industrial economy.
The Knights of Labor, founded in 1869, was the first major labor organization in the United States. The Knights organized unskilled and skilled workers, campaigned for an eight hour workday, and aspired to form a cooperative society in which laborers owned the industries in which they worked.
The Knights’ membership collapsed following the 1886 Haymarket Square riot in Chicago. By 1886 the American Federation of Labor (AFL), an alliance of skilled workers’ trade unions, was growing.
Answer:
225
Step-by-step explanation:
The sum of this series is given by the formula:
![S_n=\frac{n}{2}[2a+(n-1)d]](https://tex.z-dn.net/?f=S_n%3D%5Cfrac%7Bn%7D%7B2%7D%5B2a%2B%28n-1%29d%5D)
Where
S_n is the sum
a is the first term (we get this by plugging in n = 1 into "2n-1", we have 2(1) - 1 = 1)
n is the number of terms (the sum is defined for n = 1 to n = 15, so 15 terms)
d is the common different (the difference is successive terms. Here, 2nd term would be 2(2) - 1 = 3, and first term was 1, so d = 3 -1 = 2)
<em>plugging in the info, we will get the sum:</em>
<em>
</em>
<em>The sum is 225</em>
First we use sin(a+b)= sinacosb+sinbcosa
and cos(a+b)=cosa cosb -sinasinb
tan(x+pi/2)= sin(x+pi/2) / cos(x+pi/2)
and sin(x+pi/2) = sinxcospi/2 + sinpi/2cosx =cosx,
<span>cos(x+pi/2) = cosxcospi/2- sinxsinpi/2= - sinx,
</span> because <span>cospi/2 =0, </span>and <span>sinpi/2=1
</span><span>=tan(x+pi/2)= sin(x+pi/2) / cos(x+pi/2)= cosx / -sinx = -1/tanx = -cotx
</span>from where <span>tan(x+pi/2)=-cotx</span>
Answer:
3:6, 18:4 keep exploring man!
Step-by-step explanation:
think of numbers to make them both true like ex 3 for a and 6 for b so in this case you could do 3:6
So the question here is just simply what percentage is 10 out of 50 correct? You would basically just do 10/50 times 100 which will give you 20 percent as your answer.