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Ipatiy [6.2K]
1 year ago
13

15x²y³z÷3xy² simplify

Mathematics
2 answers:
umka21 [38]1 year ago
7 0

Answer:

5xyz

Step-by-step explanation:

\frac{15x^2y^3z}{3xy^2}

Step 1: Divide the numbers

\frac{15}{3}=5\\\\=\frac{5x^2y^3z}{xy^2}

Step 2: Simplify

\frac{x^2}{x}\\\\= x\\\\=\frac{5xy^3z}{y^2}

Step 3: Simplify

\frac{y^3}{y^2} = y\\\\=5xyz

Therefore, the simplified answer is =5xyz

posledela1 year ago
6 0

Answer:

5x^3 y^5 z

Step-by-step explanation:

See the steps below:)

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Three highways connect city A with city B. Two highways connect city B with city C.
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(a) The probability that there is no open route from A to B is (0.2)^3 = 0.008.
Therefore the probability that at least one route is open from A to B is given by: 1 - 0.008 = 0.992.
The probability that there is no open route from B to C is (0.2)^2 = 0.04.
Therefore the probability that at least one route is open from B to C is given by:
1 - 0.04 = 0.96.
The probability that at least one route is open from A to C is:
0.992\times0.96=0.9523

(b)
α The probability that at least one route is open from A to B would become 0.9984. The probability in (a) will become:0.9984\times0.96=0.95846

β The probability that at least one route is open from B to C would become 0.992. The probability in (a) will become:
0.992\times0.992=0.9841

Gamma: The probability that a highway between A and C will not be blocked in rush hour is 0.8. We need to find the probability that there is at least one route open from A to C using either a route A to B to C, or the route A to C direct. This is found by using the formula:
P(A\cup B)=P(A)+P(B)-P(A\cap B)
0.9523+0.8-(0.9523\times0.8)=0.99
Therefore building a highway direct from A to C gives the highest probability that there is at least one route open from A to C.


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2 years ago
Assume that two marbles are drawn without replacement from a box with 1 blue, 3 white, 2 green, and 2 red marbles. find the prob
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Before I answer the question i am going to start with the blue Marble you would have a 1/8 chance of drawing the blue marble from the box and after that you would have a 3/7 of drawing a white marble because you did not replace the blue marble and I hope this help.
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Which expressions are equivalent to 10a - 25 + 5b10a−25+5b10, a, minus, 25, plus, 5, b?
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Answer:

Answer: (2a - 5 + b)5

Answer: 10 x (a - 2.5 + 0.5b)

Answer: (-2a + 5 - b) ⋅ (-5)

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Camille knew that a triangle had one side with a length of 16 inches and another side with a length of 20 inches. She did not kn
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The unknown side is 9. Because 9 x 5= 45, 45 being the perimeter, that is 5 times the unknown side. And 45 is the sum of 9, 16, and 20.
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1 year 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
1 year ago
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