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4 years ago

What is the expression for the below sequence (use n) -3,-6,-9,-12

Mathematics

1 answer:

polet [3.4K]4 years ago

6 0

Answer:

$a_{n}$ = - 3n

Step-by-step explanation:

There is a common difference between consecutive terms, that is

- 6 - (- 3) = - 9 - (- 6) = - 12 - (- 9) = - 3

This indicate the sequence is arithmetic with n th term

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = - 3 and d = - 3 , thus

$a_{n}$ = - 3 - 3(n - 1) = - 3 - 3n + 3 = - 3n

The n th term is - 3n

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Read 2 more answers

7(x + y) ex2 − y2 dA, R where R is the rectangle enclosed by the lines x − y = 0, x − y = 7, x + y = 0, and x + y = 6

Anastasy [175]

Answer:

$\int\limits {\int\limits_R {7(x + y)e^{x^2 - y^2}} \, dA = \frac{1}{2}e^{42} -\frac{43}{2}$

Step-by-step explanation:

Given

$x - y = 0$

$x - y = 7$

$x + y = 0$

$x + y = 6$

Required

Evaluate $\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA$

Let:

$u=x+y$

$v =x - y$

Add both equations

$2x = u + v$

$x = \frac{u+v}{2}$

Subtract both equations

$2y = u-v$

$y = \frac{u-v}{2}$

So:

$x = \frac{u+v}{2}$

$y = \frac{u-v}{2}$

R is defined by the following boundaries:

$0 \le u \le 6$ , $0 \le v \le 7$

$u=x+y$

$\frac{du}{dx} = 1$

$\frac{du}{dy} = 1$

$v =x - y$

$\frac{dv}{dx} = 1$

$\frac{dv}{dy} = -1$

So, we can not set up Jacobian

$j(x,y) =\left[\begin{array}{cc}{\frac{du}{dx}}&{\frac{du}{dy}}\\{\frac{dv}{dx}}&{\frac{dv}{dy}}\end{array}\right]$

This gives:

$j(x,y) =\left[\begin{array}{cc}{1&1\\1&-1\end{array}\right]$

Calculate the determinant

$det\ j = 1 * -1 - 1 * -1$

$det\ j = -1-1$

$det\ j = -2$

Now the integral can be evaluated:

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA$ becomes:

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \int\limits^6_0 {\int\limits^7_0 {7ue^{x^2 - y^2}} \, *\frac{1}{|det\ j|} * dv\ du$

$x^2 - y^2 = (x + y)(x-y)$

$x^2 - y^2 = uv$

So:

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \int\limits^6_0 {\int\limits^7_0 {7ue^{uv}} *\frac{1}{|det\ j|}\, dv\ du$

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \int\limits^6_0 {\int\limits^7_0 {7ue^{uv}} *|\frac{1}{-2}|\, dv\ du$

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \int\limits^6_0 {\int\limits^7_0 {7ue^{uv}} *\frac{1}{2}\, dv\ du$

Remove constants

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2}\int\limits^6_0 {\int\limits^7_0 {ue^{uv}} \, dv\ du$

Integrate v

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2}\int\limits^6_0 \frac{1}{u} * {ue^{uv}} |\limits^7_0 du$

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2}\int\limits^6_0 e^{uv} |\limits^7_0 du$

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2}\int\limits^6_0 [e^{u*7} - e^{u*0}]du$

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2}\int\limits^6_0 [e^{7u} - 1]du$

Integrate u

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2} * [\frac{1}{7}e^{7u} - u]|\limits^6_0$

Expand

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2} * ([\frac{1}{7}e^{7*6} - 6) -(\frac{1}{7}e^{7*0} - 0)]$

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2} * ([\frac{1}{7}e^{7*6} - 6) -\frac{1}{7}]$

Open bracket

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2} * [\frac{1}{7}e^{7*6} - 6 -\frac{1}{7}]$

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2} * [\frac{1}{7}e^{7*6} -\frac{43}{7}]$

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{7}{2} * [\frac{1}{7}e^{42} -\frac{43}{7}]$

Expand

$\int\limits {\int\limits {7(x + y)e^{x^2 - y^2}} \, dA = \frac{1}{2}e^{42} -\frac{43}{2}$

3 0

3 years ago