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2 years ago

HELP PLSPLSPLS

Mathematics

1 answer:

ddd [48]2 years ago

3 0

all "atomic" or constituent statements are true

at least one "atomic" or constituent statement is true

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find the average rate of change from t=2 to t=5 for the velocity function v(t)=t^2-t+10. WILL GIVE THE BRAINLIEST ANSWER--EXPLAI

Elina [12.6K]

Answer: The average rate of change is 6.
First, plug in each value of <em>t</em> into the function, v(t) to find there coordinate pairs.

v(2) = (2)^2 - (2) + 10
v(2) = 4 + 8
v(2) = 12

v(5) = (5)^2 - (5) + 10
v(5) = 25 + 5
v(5) = 30

You can write these values as coordinate pairs, like so: (2, 12) and (5, 30).
The formula for the average rate of change is $A(x) = \frac{f(x)-(f(a)}{x-a}$ . When you plug in the values from this particular case, the average rate of change formula becomes $A(t) = \frac{30-12}{5-2}$ , or $A(t)= \frac{18}{3}$ .

Looking at the equation, you can solve for the average rate of change between t = 2 and t = 5, which equals 6.

8 0

2 years ago

Please help me! :) !!!!!!!!!!!!!!!!!

Novosadov [1.4K]

Answer:

x=17

Step-by-step explanation:

Since the triangles are congruent, the angles are equal

I≅ P

3x+4 = 72-x

Add x to each side

3x+4+x = 72-x+x

4x+4 = 72

Subtract 4 from each side

4x+4-4 = 72-4

4x= 68

Divide by 4

4x/4 = 68/4

x = 17

7 0

2 years ago

0 1 2 3 20. (03.05 MC) What does > -4 indicate about the position of - - and - 4 on the number line? (1 point) 0 2/3 is locat

liq [111]

You got this don’t give up

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2 years ago

Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (

tekilochka [14]

Answer:

Radius of convergence of power series is $\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}$

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

Power series centered at x = a is:

$\sum_{n=1}^{\infty}c_{n}(x-a)^{n}$

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

$a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]$

Applying the ratio test:

$\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}$

$\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

Applying n → ∞

$\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

The numerator as well denominator of $\frac{a_{n}}{a_{n+1}}$ are polynomials of fifth degree with leading coefficients:

$(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}$

4 0

2 years ago

Hi I need help on this question I’m struggling May someone help me pls and thank u it’s urgent!!!!

kiruha [24]

I believe you’re correct.

y = 3x - 6

8 0

2 years ago