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3 years ago

Please help me out! (show all work)

Mathematics

1 answer:

Brums [2.3K]3 years ago

3 0

First verify it holds for n=1
then you are allowed to assume it holds for n and you can start with the induction step

for n+1 instead of inserting it into the equation, add it to both sides and transform it to the original formula to proof its equality

$1^{2}+2^{2}+3^{2}+...+n^{2}+(n+1)^{2}=\frac{n(n+1)(2n+1)}{6} +(n+1)^{2} \\ =\frac{n(n+1)(2n+1)+6(n+1)^{2} }{6}\\ =\frac{n(n+1)(2n+1)+6(n+1)(n+1)}{6}\\ =\frac{(n+1)(n(2n+1)+6(n+1))}{6}\\ =\frac{(n+1)((2n^{2}+n)+6(n+1))}{6}\\ =\frac{(n+1)(2n^{2}+7n+6)}{6}\\ =\frac{(n+1)(n+2)(2n+3)}{6}\\ =\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$

this equals the original formula if you had inserted n+1, proving the correctness

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In a baseball league consisting of 15 teams, each team plays each of the other teams three times. How many games will be played

denis23 [38]

Answer:

315

Step-by-step explanation:

Each of the 15 teams plays each of the other 14, for a total of 210 games. However, that count includes both A playing B and B playing A. Hence, if the teams meet 3 times, the total number of games will be 3/2 this amount:

(15)(14)(3/2) = 315 . . . league games

5 0

3 years ago

A function f is _____ on an open interval I if, for any choice of x1 and x2 in I, with 1 less than

Zepler [3.9K]

Answer:

If $f(x_1)\leq f(x_2)$ whenever $x_1\leq x_2$ f is increasing on I.

If $f(x_1)\geq f(x_2)$ whenever $x_1\leq x_2$ f is decreasing on I.

Step-by-step explanation:

These are definitions for real-valued functions f:I→R. To help you remember the definitions, you can interpret them in the following way:

When you choose any two numbers $x_1\leq x_2$ on I and compare their image under f, the following can happen.

$f(x_1)\leq f(x_2)$ . Because x2 is bigger than x1, you can think of f also becoming bigger, that is, f is increasing. The bigger the number x2, the bigger f becomes.

$f(x_1)\geq f(x_2)$ . The bigger the number x2, the smaller f becomes so f is "going down", that is, f is decreasing.

Note that this must hold for ALL choices of x1, x2. There exist many functions that are neither increasing nor decreasing, but usually some definition applies for continuous functions on a small enough interval I.

3 0

3 years ago

Which is the value of the expression (StartFraction (10 Superscript 4 Baseline) (5 squared) Over (10 cubed) (5 cubed)) cubed?

Flura [38]

Answer:

The value to the given expression is 8

Therefore $\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=8$

Step-by-step explanation:

Given expression is (StartFraction (10 Superscript 4 Baseline) (5 squared) Over (10 cubed) (5 cubed)) cubed

Given expression can be written as below

$\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3$

To find the value of the given expression:

$\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=\frac{((10^4)(5^2))^3}{((10^3)(5^3))^3}$

( By using the property ( $(\frac{a}{b})^m=\frac{a^m}{b^m}$ )

$=\frac{(10^4)^3(5^2)^3}{(10^3)^3(5^3)^3}$

( By using the property $(ab)^m=a^mb^m$ )

$=\frac{(10^{12})(5^6)}{(10^9)(5^9)}$

( By using the property $(a^m)^n=a^{mn}$ )

$=(10^{12})(5^6)(10^{-9})(5^{-9})$

( By using the property $\frac{1}{a^m}=a^{-m}$ )

$=(10^{12-9})(5^{6-9})$ (By using the property $a^m.b^n=a^{m+n}$ )

$=(10^3)(5^{-3})$

$=\frac{10^3}{5^3}$ ( By using the property $a^{-m}=\frac{1}{a^m}$ )

$=\frac{1000}{125}$

$=8$

Therefore $\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=8$

Therefore the value to the given expression is 8

3 0

3 years ago

Read 2 more answers

Which of these is the statement of zero product rule?

Jobisdone [24]

Well there are no statements but the zero power law works like this :
ANYTHING raised to the zero power is 1
that is anything BUT zero itself because 0 to the power of 0vis still 0 and it makes no sense to write 0 to the power of zero.

3 0

3 years ago

Find all roots: x^3 + 7x^2 + 12x = 0 Show all work and check your answer.

Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

Solution:

We have been given a cubic polynomial.

$x^{3}+7 x^{2}+12 x=0$