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2 years ago

A triangle has side lengths of 5 inches, 8 inches, and 9 inches. What are possible side lengths of a similar triangle?

1 answer:

3 0

The answer is A: 30 in, 48 in, and 54 in.

Explanation:
If you multiply all of the sides by 6 you get answer. 5 * 6 = 30, 8 * 6 = 48, and
9 * 6 = 54.

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A lender wil verity and carefully consider your income before approving you for a loan because

ExtremeBDS [4]

A. They need to be sure you will be able to pay the loan back

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2 years ago

Read 2 more answers

Restless Leg Syndrome and Fibromyalgia

Anna11 [10]

Answer:

The probability that a person with restless leg syndrome has fibromyalgia is 0.183.

Step-by-step explanation:

Denote the events as follows:

<em>F</em> = a person with fibromyalgia

<em>R</em> = a person having restless leg syndrome

The information provided is as follows:

P (R | F) = 0.33

P (R | F') = 0.03

P (F) = 0.02

Consider the tree diagram attached below.

Compute the probability that a person with restless leg syndrome has fibromyalgia as follows:

$P(F|R)=\frac{P(R|F)P(F)}{P(R|F)P(F)+P(R|F')P(F')}$

$=\frac{(0.33\times 0.02)}{(0.33\times 0.02)+(0.03\times 0.98)}\\\\=\frac{0.0066}{0.0066+0.0294}\\\\=0.183333\\\\\approx 0.183$

Thus, the probability that a person with restless leg syndrome has fibromyalgia is 0.183.

3 0

2 years ago

I feed to find the circumference

fgiga [73]

answer:

131.88 inches

step-by-step explanation:

know the formula: 2πr
now plug in what we have

2πr = circumference

2(21)(3.14)

= 131.88

6 0

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

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

7 0

2 years ago

I have to do Decompose fractions and my problem is 3 over 7 what do I do?

mart [117]

Just simplify them to a decimal by dividing the numerator by the denominator.

4 0

2 years ago