4 cos² x - 3 = 0
4 cos² x = 3
cos² x = 3/4
cos x = ±(√3)/2
Fixing the squared cosine doesn't discriminate among quadrants. There's one in every quadrant
cos x = ± cos(π/6)
Let's do plus first. In general, cos x = cos a has solutions x = ±a + 2πk integer k
cos x = cos(π/6)
x = ±π/6 + 2πk
Minus next.
cos x = -cos(π/6)
cos x = cos(π - π/6)
cos x = cos(5π/6)
x = ±5π/6 + 2πk
We'll write all our solutions as
x = { -5π/6, -π/6, π/6, 5π/6 } + 2πk integer k
Answer:
HI
Step-by-step explanation:
Yes, because a rational number is any number that can be expressed as a fraction a/b where a/b are both integers, but cannot be zero.
Answer:
This sampling method used in this question is the stratified sampling method.
Step-by-step explanation:
There are about 5 known sampling methods.
- Random Sampling
In random sampling, each member of the population has an equal chance of being surveyed. All the students are given a number and random numbers are generated to pick the students to be surveyed.
- Systematic sampling
This is easier than random sampling. In systematic sampling, a particular number, n, is counted repeatedly and each of the nth student is picked to be sampled.
- Convenience Sampling
This is the worst sampling technique. It is also the easiest. In Convenience sampling, the surveyor just surveys the first set of students that they find.
- Cluster sampling
Cluster Sampling divides the population into groups which are called clusters or blocks. The clusters are selected randomly, and every element in the selected clusters is surveyed.
- Stratified Sampling
Stratified sampling divides the population into groups called strata. A sample is taken from all or some of these strata using either random, systematic, or convenience sampling. This is evidently the answer to the question as the students are divided into homerooms (strata) and samples are now randomly taken from 3 randomly selected strata.
Hope this Helps!!!
Answer:
a) 820
b)450
c) -540
d) -294
e) 1440
f) 1425
Step-by-step explanation:
a) it is an arithmetic progression with ratio=4
a1=3,
a2=3+r=3+4=7.....
a20=a1+19r
so a20=3+19*4=3+76=79
S=(a1+an)*n/2 where n=20
so S=(3+79)*20/2=820
b) a15=a1+14*r=2+14*4=2+56=58
S=(a1+a15)*15/2=(2+58)*15/2=60*15/2=30*15=450
It is the same formula for all the exercises
S=(a1+an)*n/2 , where n is the number of terms
c) a40=30+39*(-3)=30-87=-57
So S=(30-57)*40/2=-540
d) a14=5+13*(-4)=5-52=-47
S=(5-47)*14/2=-42*14/2=-42*7=-294
e) it is 5+7+9+......+75
nr of terms = (an-a1) :r+1=(75-5):2+1=36
S=(5+75)*36/2=36*40=1440
f) ratio=7-4=3
the same formula for n= (91-4):3+1=87:3+1=29+1=30
S=(4+91)*30/2=95*30/2=2850/2=1425
