What is the common difference of the following sequence?3,5,7,9,11_16,19,22,25_

Mathematics

1 answer:

valina [46]3 years ago

3 0

Answer:

the first one goes by 4n-1

the Second sequence goes by 3n + 13

Step-by-step explanation:

using the formula an=a1+d(n-1)

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Please please help!!

shepuryov [24]

Answer:

$\large\boxed{x=0\ and\ x=\pi}$

Step-by-step explanation:

$\tan^2x\sec^2x+2\sec^2x-\tan^2x=2\\\\\text{Use}\ \tan x=\dfrac{\sin x}{\cos x},\ \sec x=\dfrac{1}{\cos x}:\\\\\left(\dfrac{\sin x}{\cos x}\right)^2\left(\dfrac{1}{\cos x}\right)^2+2\left(\dfrac{1}{\cos x}\right)^2-\left(\dfrac{\sin x}{\cos x}\right)^2=2\\\\\left(\dfrac{\sin^2x}{\cos^2x}\right)\left(\dfrac{1}{\cos^2x}\right)+\dfrac{2}{\cos^2x}-\dfrac{\sin^2x}{\cos^2x}=2$

$\dfrac{\sin^2x}{(\cos^2x)^2}+\dfrac{2-\sin^2x}{\cos^2x}=2\\\\\text{Use}\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{2-(1-\cos^2x)}{\cos^2x}=2\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{2-1+\cos^2x}{\cos^2x}=2\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{1+\cos^2x}{\cos^2x}=2$

$\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{(1+\cos^2x)(\cos^2x)}{(\cos^2x)^2}=2\qquad\text{Use the distributive property}\\\\\dfrac{1-\cos^2x+\cos^2x+\cos^4x}{\cos^4x}=2\\\\\dfrac{1+\cos^4x}{\cos^4x}=2\qquad\text{multiply both sides by}\ \cos^4x\neq0\\\\1+\cos^4x=2\cos^4x\qquad\text{subtract}\ \cos^4x\ \text{from both sides}\\\\1=\cos^4x\iff \cos x=\pm\sqrt1\to\cos x=\pm1\\\\ x=k\pi\ for\ k\in\mathbb{Z}\\\\\text{On the interval}\ 0\leq x$

4 0

3 years ago

Find the distance between<br> 2-3i<br> And<br> 9+21i

Goryan [66]

2 - 3i is just another of writing (2, -3) in the cartesian plane, the 2-3i however is using the "imaginary axis" for the imaginary plane, is all however the imaginary axis is simply the equivalent of the y-axis. Likewise 9+21i is just (9,21), so let's just use the distance formula.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{9}~,~\stackrel{y_2}{21})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[9-2]^2+[21-(-3)]^2}\implies d=\sqrt{(9-2)^2+(21+3)^2} \\\\\\ d=\sqrt{7^2+24^2}\implies d=\sqrt{49+576}\implies d=\sqrt{625}\implies d=25$