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2 years ago

Answer for brainlist and more points ☺️

Mathematics

1 answer:

nexus9112 [7]2 years ago

6 0

Answer:

a b and f

Step-by-step explanation:

it is what it issssssss

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HELP HELP HELP HELP HELP HELP AHHLIDAKUYSWJG PLEASE

Gelneren [198K]

I'm pretty sure it's D

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2 years ago

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Colin’s piano lesson is 45 minutes, or Three-fourths of an hour, long. If he practices 6 different songs and spends an equal amo

NikAS [45]

Answer:

<em>Colin spends 1/8 of an hour practicing each song</em>

Step-by-step explanation:

Colin's piano lesson is 3/4 of an hour, long.

In that period, he practices 6 different songs, spending an equal amount of time on each one.

This means that each song gets the sixth part of the lesson duration. The fraction that expresses that amount is:

$\displaystyle \frac{\frac{3}{4}}{6}=\frac{3}{24}=\frac{1}{8}$

Colin spends 1/8 of an hour practicing each song

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2 years ago

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emmasim [6.3K]

Answer:

Not equivalent

Step-by-step explanation:

Distribute the negative in the parentheses. It basically multiplying everything inside by -1.

$( \frac{1}{2} + 6x + 3)$

add like terms

$6x + 3 \frac{1}{2}$

Not equivalent since the 6x is positive

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2 years ago

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Are the expressions equivalent? 10(x + 2); 10x + 5

lutik1710 [3]

Answer: yes, indeed it is i forgot how it is but it is

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2 years ago

1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING

iogann1982 [59]

$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

then

$y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$

The ODE in terms of these series is

$\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0$

$\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0$

$\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}$

We can solve the recurrence exactly by substitution:

$a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0$

$\implies a_n=\dfrac{(-2)^n}{n!}a_0$

So the ODE has solution

$y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}$

which you may recognize as the power series of the exponential function. Then

$\boxed{y(x)=a_0e^{-2x}}$

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2 years ago