Answer:
is equivalent to the inequality 
Step-by-step explanation:
Given inequality :
We are supposed to find Which of the following is equivalent to the inequality
Multiply both sides by 10
is equivalent to the inequality
So, Option B is true
B)
The answer to your question is C
Answer:
L.S = R.S ⇒ Proved down
Step-by-step explanation:
Let us revise some rules in trigonometry
- sin²α + cos²α = 1
- sin2α = 2 sin α cosα
- cscα = 1/sinα
To solve the question let us find the simplest form of the right side and the left side, then show that they are equal
∵ L.S = csc2α + 1
→ By using the 3rd rule above
∴ L.S = 
+ 1
→ Change 1 to 
∴ L.S = +
→ The denominators are equal, then add the numerators
∴ L.S = 
∵ R. S = 
∵ (sinα + cosα)² = sin²α + 2 sinα cosα + cos²α
∴ (sinα + cosα)² = sin²α + cos²α + 2 sinα cosα
→ By using the 1st rule above, equate sin²α + cos²α by 1
∴ (sinα + cosα)² = 1 + 2 sinα cosα
→ By using the 2nd rule above, equate 2 sinα cosα by sin2α
∴ (sinα + cosα)² = 1 + sin2α
→ Substitute it in the R.S above
∴ R. S =
∵ L.S = R.S
∴ csc 2α + 1 =
Answer:
3 -1 -2
5 1 6
Step-by-step explanation:
An augmented system has the coefficients for the variables and then the solution going across
Rewriting the equations to get them in the form
ax + by = c
-3x+y =2
3x-y =-2
5x+y = 6
The matrix is
3 -1 -2
5 1 6
<span>(a) At the end of Month 0, about how many more insects were in Pod A than Pod B? Explain.
In Pod A, the point is higher than 50, it could be 60 to 70 insects. Pod B has 20 insects. So, Pod A has at least 40 insects more than Pod B.
(b) Find and compare the growth rates of each pod. Show your work.
Pod A: (0,60) ; (1,80) ; (2,100)
(80-60)/60 = 0.33
(100-80)/80 = 0.25
Pod B: (0,20) ; (1,44) ; (2,97)
(44-20)/20 = 1.2
(97-44)/44 = 1.2
Based on my computation, the rate of Pod A is lower than the rate of Pod B.
(c) When does the population in Pod B exceed the population in Pod A? Explain.
Pob B exceeds the population of Pod A at the END OF MONTH 4.
Pod A has a population of less than 200 while Pod B has a population of 469.</span>
