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2 years ago

Given that MQL =180° and XQR = 180°, which equation could be used to solve problems involving the relationships

Mathematics

1 answer:

Ganezh [65]2 years ago

8 0

Answer:

Option A is the right answer.

146-4b = 122-1b

Reason: vertical angles

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Wally purchased a desk that was on sale do 2/3 of the original price. If the original price was $450, what was the price that Wa

Arisa [49]

Answer:

2/3 of 450 is 300 :)

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2 years ago

Which fraction is greater than 4/8?

Ira Lisetskai [31]

3/4
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3 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

Find x...............................

notsponge [240]

Answer:

sqrt(2) =x

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

sin theta = opp/ hyp

sin 45 = x/2

2 sin 45 =x

2 ( sqrt(2)/2) =x

sqrt(2) =x

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2 years ago

Read 2 more answers

Teresa has 3 pairs of dress shoes and 4 pairs of sneakers. Which describes a ratio of 7 : 3? A Total pairs to dress pairs B Pair

denis23 [38]

A. Total pairs to dress pairs

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2 years ago