Answer:
See Below.
Step-by-step explanation:
Problem A)
We have:
When in doubt, convert all reciprocal trig functions and tangent into terms of sine and cosine.
So, let cscθ = 1/sinθ and tanθ = sinθ/cosθ. Hence:
Cancel:
Let 1/cosθ = secθ:
From the Pythagorean Identity, we know that tan²θ + 1 = sec²θ. Hence, sec²θ - 1 = tan²θ:
Problem B)
We have:
Factor out a sine:
From the Pythagorean Identity, sin²θ + cos²θ = 1. Hence, sin²θ = 1 - cos²θ:
Distribute:
Problem C)
We have:
Recall that cos2θ = cos²θ - sin²θ and that sin2θ = 2sinθcosθ. Hence:
From the Pythagorean Identity, sin²θ + cos²θ = 1 so cos²θ = 1 - sin²θ:
Cancel:
By definition:
Not an expertise on infinite sums but the most straightforward explanation is that infinity isn't a number.
Let's see if there are anything we missed:
∞
Σ 2^n=1+2+4+8+16+...
n=0
We multiply (2-1) on both sides:
∞
(2-1) Σ 2^n=(2-1)1+2+4+8+16+...
n=0
And we expand;
∞
Σ 2^n=(2+4+8+16+32+...)-(1+2+4+8+16+...)
n=0
But now, imagine that the expression 1+2+4+8+16+... have the last term of 2^n, where n is infinity, then the expression of 2+4+8+16+32+... must have the last term of 2(2^n), then if we cancel out the term, we are still missing one more term to write:
∞
Σ 2^n=-1+2(2^n)
n=0
If n is infinity, then 2^n must also be infinity. So technically, this goes back to infinity.
Although we set a finite term for both expressions, the further we list the terms, they will sooner or later approach infinity.
Yep, this shows how weird the infinity sign is.
The correct answer is:
D) Jeffery's conclusion is not valid because the sample was biased since all of the students were eighth grade students.
Answer:
-4x+y = -5
Step-by-step explanation:
Y+7=4(x-3)
Y+7=4x-12
+7. +7
Y=4x-5
-4x+y=-5
