All categories

2 years ago

All proofs start as ______. A.) Constructions B.) Conjectures C.) Converses

Mathematics

2 answers:

trasher [3.6K]2 years ago

8 0

I’m pretty sure it’s c

Valentin [98]2 years ago

7 0

Answer: Conjectures.

Step-by-step explanation: A conjecture is an opinion or conclusion formed on the basis of incomplete information. All proofs start as a conjecture until it is proven as a truth (a proof).

You might be interested in

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent p-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

5 0

2 years ago

The perimeter of a triangle is 104 units. The combined length of two of the sides of the triangle is 64 units.

valkas [14]

Answer:

Step-by-step explanation:

168

7 0

1 year ago

Seth's balcony is 7 feet wide and 14 feet long. Seth paid $4.00 per square foot to have the

BlackZzzverrR [31]

Answer:

392

Step-by-step explanation:

4 0

2 years ago

Please help! show explanation please

Rudiy27

Hey use photo math is gonna help you a lot

4 0

2 years ago

The time to complete a standardized exam is approximately normal with a mean of 70 minutes and a standard deviation of 10 minus.

AnnyKZ [126]

Answer:

Step-by-step explanation:

Given that the time to complete a standardized exam is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes.

P(completing exam before 1 hour)

= P(less than an hour) = P(X<60)

=P(Z< $\frac{60-70}{10} =-1$ )

=0.5-0.34=0.16

i.e. 16% of students completed the standardized exam.

7 0

3 years ago