Answer:
John should use a protractor to measure an angle of 70°.
Step-by-step explanation:
70° is an angle for which there is no construction procedure.
Angles of measures 15°, 18°, 30°, 36°, and 45° can be constructed, along with their supplements, complements, and values multiplied by any positive integer or divided by any power of 2. Of course, the sums and differences of any of these angles can be constructed. (The more steps in the construction, the greater the loss of accuracy.)
Because there is no construction procedure for an angle of 70°, John must create the angle using a measuring tool. A protractor is the tool of choice.
John should use a protractor to measure an angle of 70° relative to one end of his 5 cm baseline.
_____
<em>Comment on John's technique</em>
It is likely John would create a more accurate representation of the desired triangle by drawing the 7 cm segment first. This would give a longer baseline for use with the protractor, and would mean the length through the point marked using the protractor would not have to be extrapolated as far.
Answer:
3
Step-by-step explanation:
Answer:
The first, second, and fifth statements are correct.
Step-by-step explanation:
We are given a circle with the equation:

And we want to select the statements that are true.
First, we can convert the equation into standard form. We can group each variable:

And complete the square for the first term:

Factor and simplify:

We can rewrite our equation as:

So, this tells us that we have a circle centered on (1, 0) with a radius of 3 units.
In this case, the first statement, second statement (the point (1,0) is on the x-axis), and fifth statements are correct (the square root of 9 is also 3).
Mickey mouse and donald duck
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.