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4 years ago

How do you know when a sequence is arithmetic vs. geometric?

Mathematics

1 answer:

stellarik [79]4 years ago

8 0

Answer:

An arithmetic sequence is a sequence with the difference or pattern between two consecutive terms constant.

A geometric sequence is a sequence with a ratio between two consecutive terms constant.

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−17+17+5=(−17+17)+5=

nika2105 [10]

Answer:

-17+17=0

0+5=5

-17+17=0

5 0

3 years ago

Read 2 more answers

What is x, if the volume of the cylinder is 768pi rcm3?

AVprozaik [17]

let's recall Cavalieri's Principle, <u>solids with equal altitudes and cross-sectional areas at each height have the same volume</u>, so even though this cylinder is slanted with a height = x and a radius = 8, the cross-sectional areas from the bottom to top are the same thickness and thus the same area, so its volume will be the same as a cylinder with the same height and radius that is not slanted.

$\bf \textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\ \cline{1-1} r=8\\ h=x\\ V=768\pi \end{cases}\implies 768\pi =\pi (8)^2(x)\implies 768\pi =64\pi x \\\\\\ \cfrac{768\pi }{64\pi }=x\implies 12=x$

5 0

3 years ago

Three desks and a bookcase together weigh 157Ib the bookcase weighs 34Ib how much does each desk weigh

Fantom [35]

Desk weight = d
Bookcase weight = b
b = 34
3d + b = 157
3d + 34 = 157
D = 41

5 0

3 years ago

What's the greatest common number between 5,11

Katen [24]

11 the greatest common number between 5,11

3 0

4 years ago

In the right ΔABC, AN is the altitude to the hypotenuse. Find BN, AN, and AC, if AB=2√5 in and NC= 1 in.

kipiarov [429]

Answer:

$BN=4\ \text{in}$

$AN=2\ \text{in}$

$AC=\sqrt{5}\ \text{in}$

Step-by-step explanation:

Let $BN=x$

$BC=c$

$NC=y=1\ \text{in}$

$AB=a=2\sqrt{5}\ \text{in}$

$AC=b$

We have the relation

$\dfrac{BC}{AB}=\dfrac{AB}{BN}\\\Rightarrow \dfrac{c}{a}=\dfrac{a}{x}\\\Rightarrow c=\dfrac{a^2}{x}\\\Rightarrow x+1=\dfrac{a^2}{x}\\\Rightarrow x^2+x=20\\\Rightarrow x^2+x-20=0\\\Rightarrow x=\frac{-1\pm \sqrt{1^2-4\times 1\times \left(-20\right)}}{2\times 1}\\\Rightarrow x=4,-5$

$\boldsymbol{BN=4\ \text{in}}$

$h=\sqrt{a^2-x^2}\\\Rightarrow h=\sqrt{(2\sqrt{5})^2-4^2}\\\Rightarrow h=2\ \text{in}$

$\boldsymbol{AN=2\ \text{in}}$

$b=\sqrt{h^2+y^2}\\\Rightarrow b=\sqrt{2^2+1^2}\\\Rightarrow b=\sqrt{5}\ \text{in}$

$\boldsymbol{AC=\sqrt{5}\ \text{in}}$

8 0

3 years ago