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skad [1K]
2 years ago
10

The first 5 terms of a quadratic sequence are

Mathematics
1 answer:
Mashcka [7]2 years ago
8 0

Step-by-step explanation:

an = 2n^2 - 3 · Anonymous. -1, 5, 15, 29, 47, ... an = 2 n^2 - 3.

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What property is shown in 3x+4y+2+6y+3
True [87]

Answer:

Step-by-step explanation:

I think its substitution

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2 years ago
7 over 14 in its simplest form
eduard

Answer:

1/2

Step-by-step explanation:

Half of 14 is 7 therefore it is 1/2

3 0
3 years ago
At the farmer’s market, you can buy 3 melons for $13.50. What is the cost of one melon
Virty [35]

Answer:

4.50 per melon

Step-by-step explanation:

Take the price and divide by the number of melons

13.50/3

4.50 per melon

7 0
3 years ago
Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o
koban [17]

Answer:  17.6 grams

Step-by-step explanation:

As the problem tells us, the velocity of the reaction is proportional to the product of the quantities of A and B that have not reacted, so from this we get the next equation:

                                                       V = k[A][B]

where [A] represents the remaining amount of A, and [B] represents the remaining amount of B. To solve this equation we have to represent it through a differential equation, which is:

                                              dx/dt = k[α - a(t)][β - b(t)]         (1)

where,

k: velocity constant

a(t): quantity of A consumed in instant t

b(t): quantity of B consumed in instant t

α: initial quantity of A

β: initial quantity of B

Now we need to define the equations for a(t) and b(t), and for this we are going to use the law of conservation of mass by Lavoisier, with which we can say that the quantity of C in a certain instant is equal to the sum of the quantities of A and B that have reacted. Therefore, if we need M grams of A and N grams of B to form a quantity of M+N of C, then we can say that in a certain time, the consumed quantities of A and B are given by the following equations:

                                       a(t) = ( M/M+N) · x(t)

                                       b(t) = (N/M+N) · x(t)

where,

x(t): quantity of C in instant t

So for this problem we have that for 1 gram of B, 2 grams of A are used, therefore the previous equations can be represented as:

                                       a(t) = (2/2+1) · x(t) = 2/3 x(t)

                                       b(t) = (1/2+1) · x(t) = 1/3 x(t)

Now we proceed to resolve the differential equation (1) by substituting values:

                                         dx/dt = k[α - a(t)][β - b(t)]  

                                        dx/dt = k[40 - 2x/3][50 - x/3]

                                         dx/dt = k/9 [120 - 2x][150 - x]

We use the separation of variables method:

                                      dx/[120-2x][150-x] = k/3 · dt

We integrate both sides of the equation:

                                     ∫dx/(120-2x)(150-x) = ∫kdt/9

                                     ∫dx/(15-x)(60-x) = kt/9 + c

Now, to integrate the left side of the equation we need to use the partial fraction decomposition:

                                    ∫[1/90(120-2x) - 1/180(150-x)] = kt/9 + c

                                      1/180 ln(150-x/120-2x) = kt/9 + c

                                           (150-x)/(120-2x) = Ce^{20kt}

Now we resolve by taking into account that x(0) = 0, and x(5) = 10,

for x(0) = 0 ,             (150-0)/(120-0) = Ce^{20k(0)} , C = 1.25

for x(5) = 10 ,           (150-10)/(120-(2·10)) = 1.25e^{20k(5)} , k ≈ 113 · 10^{-5}

Now that we have the values of C and k, we have this equation:

                           (150-x)/(120-2x) = 1.25e^{226·10^{-4}t}

and we have to clear by x, obtaining:

               x(t) = 150 · (1 - e^{226·10^{-4}t} / 1 - 2.5e^{226·10^{-4}t})

Therefore the quantity of C that will be formed in 10 minutes is:

           x(10) = 150 · (1 - e^{226·10^{-4}(10)} / 1 - 2.5e^{226·10^{-4}(10)})

                                            x(10) ≈ 17.6 grams

8 0
3 years ago
A normal distribution has a mean of µ = 100 with σ = 20. if one score is randomly selected from this distribution, what is the p
avanturin [10]

Given: Mean of Normal distribution μ = 100, standard deviation σ =20

We have to find the probability that the score will be less than X=84

P(X < 84)

The z-score for x=84 is

Z = \frac{x-mean}{standard deviation}

Z = \frac{84-100}{20}

Z = -0.8

We have to find probability that Z < -0.8

Using Z score table to find the probability below z=-0.8 we get

P(Z < -0.8) = 0.2119

Hence the probability that the score will be less than x = 84 is 0.2119

8 0
3 years ago
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