Help‼️ if a number is even, then it is divisible by 2. g=14

WILL GIVE BRAINLIEST-What is the average rate of change of the function on the interval from x = 0 to x = 2?

Musya8 [376]

Average= f(2)−f(0)/2−0
=62.5−250/ 2-0
= −93.75

7 0

3 years ago

Match the types of well formed formulas with the expressions. Negation Conjunction Disjunction Conditional Biconditional A. T &l

nikklg [1K]

<h3>Answers:</h3>

A. T <-> U is a biconditional
B. (A & B) v (C & D) is a disjunction
C. R -> ~S is a conditional
D. P & Q is a conjunction
E. ~(R v P) is a negation

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Explanations:

A biconditional is anything in the form A <-> B. This is a compact way of saying (A -> B) & (B -> A). We replace A and B with logical statements.
Disjunctions are of the basic form A v B. The "v" basically means "or".
Any conditional is of the form "if... then...". For example, "if it rains, then it gets wet outside" is a conditional. In terms of logic symbols, we write A -> B to mean "if A, then B".
Conjunctions are whenever we combine two logical statements with an "and" or an ampersand symbol. The basic form is A & B
Negations are the complete opposite of the original. If the original is P, then the negation is ~P, which is read as "not P".

6 0

4 years ago

Can someone please help me out ?

gizmo_the_mogwai [7]

Answer:

205-160=d

Step-by-step explanation:

4 0

4 years ago

Read 2 more answers

Which of the following happens when 21.63 is multiplied by 103?

anastassius [24]

Answer: A
Because it is

7 0

4 years ago

Consider the function f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0

dybincka [34]

I suppose you mean

$f(x)=\cos(x^2)$

Recall that

$\cos x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}$

which converges everywhere. Then by substitution,

$\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}$

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

$f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}$

(starting at $n=1$ because the summand is 0 when $n=0$ )

b. Naturally, the differentiated series represents

$f'(x)=-2x\sin(x^2)$

To see this, recalling the series for $\sin x$ , we know

$\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}$

Multiplying by $-2x$ gives

$-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}$

and from here,

$-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}$

$-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)$

c. This series also converges everywhere. By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}$

The limit is 0, so any choice of $x$ satisfies the convergence condition.

3 0

4 years ago