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Advocard [28]
3 years ago
6

Which is a line segment?a. 'bar(PQ)'b. 'bar(RS)'c. 'bar(TU)'d. 'bar(VW)'

Mathematics
1 answer:
Julli [10]3 years ago
8 0
B. 'bar(RS)
because a line segment has 2 dots at the end and the beginning nothing goes over the 2 dots
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The integer -3 would BEST represent which of these events?
Levart [38]
The correct answer would be number 3 only because you are taking away when you have a negative number. SO u are taking away 3 gallons of gas. 
Hope this helped!
5 0
3 years ago
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Nine students in Maddie's classroom are wearing red. There are
goblinko [34]

Answer:

She is correct because 30 percent of 30 is 9 or in other words if you multiply 30 by 30 percent you get 9.

Step-by-step explanation:

5 0
3 years ago
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As a decimal 6 27/100​
Brut [27]

Answer:

6.27

Step-by-step explanation:

627100=6+27100

We know that

27100

is the same as

27÷100

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8 0
4 years ago
Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (
musickatia [10]
(a) Suppose h_n=r^n is a solution for this recurrence, with r\neq0. Then

r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}
\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}
\implies r^4-5r^3+6r^2+4r-8=0
\implies (r-2)^3(r+1)=0\implies r=2,r=-1

So we expect a general solution of the form

h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n

With h_0=0,h_1=1,h_2=1,h_3=2, we get four equations in four unknowns:

\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}

So the particular solution to the recurrence is

h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n

(b) Let G(x)=\displaystyle\sum_{n\ge0}h_nx^n be the generating function for h_n. Multiply both sides of the recurrence by x^n and sum over all n\ge4.

\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n
\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n
G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)
G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)
(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3
G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}
G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of x^n, ideally matching the solution found in part (a).
3 0
3 years ago
BRAINLIEST IF CORRECT PLEASE SHOW WORK:)
Licemer1 [7]
First of all, you will want to change the fraction to have the same denominator.

The common factor for 9 and 6 is 18. (9x2=18 and 6x3=18)
Then, you multiply the numerator by the factor you got in the first step, which is 2 and 3, respectively.
So the new fractions are \frac{4}{18} and 7 \frac{3}{18}

The next step is to simply subtract the mixed numbers!
8 \frac{4}{18} - \frac{3}{18} = 1 1/18

The answer would be <span>D: 1 1/18!</span>
4 0
3 years ago
Read 2 more answers
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