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Korolek [52]
2 years ago
14

The first three terms of a sequence are 6, 12, 30, . . .

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

Answer:

306

Step-by-step explanation:

Take a list of primes and multiply each with the prime+1:

2*3 = 6

3*4 = 12

5*6 = 30

7*8 = 56

11*12 = 132

13*14 = 182

17*18 = 306

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vladimir2022 [97]
Either one at any given time could be equal to zero. So
x + 2 = 0
x = - 2

x - 18 = 0
x = 18

I've enclosed a graph of this so you can see that the answers I've given are the ones expected.

6 0
3 years ago
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hodyreva [135]
This is the answer...


y=250-25x
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5 0
3 years ago
During the mayoral election,two debates were held between the candidates. The first debate lasted 1 2/3 hours.The second one was
butalik [34]

Answer:

2nd debate was 3 hours long.

Step-by-step explanation:

We have been given that during the mayoral election,two debates were held between the candidates. The first debate lasted 1 2/3 hours. The second one was 1 4/5 times as long as the first one.

Let us find the estimate of time spent on 2nd debate.

1 2/3 hours would be approximately 2 hours. 1 4/5 times would be equal to 2 times.

2*2=4

Therefore, the estimated time is less 4 hours.

To find the time spent on 2nd debate, we will multiply 1 2/3 by 1 4/5.

First of all, we will convert mixed fractions into improper fractions as:

1\frac{2}{3}=\frac{5}{3}

1\frac{4}{5}=\frac{9}{5}

Now, we will multiply both fractions as:

\frac{5}{3}\times \frac{9}{5}

\frac{1}{3}\times \frac{9}{1}

\frac{9}{3}\Rightarrow 3

Therefore, the 2nd debate was 3 hours long.

5 0
3 years ago
Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.
fredd [130]

\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}

\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}

so the ODE is indeed exact and there is a solution of the form F(x,y)=C. We have

\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)

\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)

f'(y)=0\implies f(y)=C

\implies F(x,y)=x^2y^3+\ln(x+y^2)=C

With y(1)=2, we have

8+\ln9=C

so

\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}

8 0
2 years ago
The measure of angle 1 is 3k - 7 and the measure of angle 8 is 5k + 29 29. What is the value of k
Fofino [41]
D: k= 36 it is that one because it’s a big number
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