All categories

4 years ago

Consider the first four terms of an arithmetic sequence below.

Mathematics

1 answer:

VMariaS [17]4 years ago

6 0

7, 13, 19 and 25 have a common difference: 6.
6 added to 7 gives us 13; 6 added to 13 gives us 19, and so on.

Explicit formula: a(n) = 7 + 6(n-1), where 7 is the first term and n is the counter (1, 2, 3, ...).

The first term is 7 (given). This corresponds to n=1.

The second term is a(2) = 7 + 6(2-1), or 7 + 6, or 13. This corresponds to n = 2.

and so on.

You might be interested in

How can I write 0.4 in word form ?

Art [367]

Answer:

four tenths

Step-by-step explanation:

5 0

3 years ago

How do you round 3,284,994 to the nearest ten thousand

Ghella [55]

3,290,000 is the answer

6 0

3 years ago

Read 2 more answers

In which quadrants could Zach graph points? Explain your thinking.

irga5000 [103]

Quadrants are only quadrant 1 and 4

8 0

3 years ago

Which exponential function repreasents growth?

4vir4ik [10]

Answer:

There are two types of exponential functions: exponential growth and exponential decay. In the function f (x) = bx when b > 1, the function represents exponential growth.

Step-by-step explanation: Math websites man

8 0

2 years ago

A curve is traced by a point P(x, y) which moves such that its distance from the point A(-1,1) is three times its distance from

docker41 [41]

Answer:

$8x^2+8y^2+43-38x+20y=0$

Step-by-step explanation:

Let $A\left ( x_1,y_1 \right )$ and $B\left ( x_2,y_2 \right )$ be two points then distance AB is equal to $AB=\sqrt{\left ( x_2-x_1 \right )^2+\left ( y_2-y_1 \right )^2}$

Here, a curve is traced by a point $P(x,y)$ which moves such that its distance from the point $A(-1,1)$ is three times its distance from the point $B(2,-1)$ i.e $AP=3BP$

Using distance formula,

$AP=\sqrt{(-1-x)^2+(1-y)^2}$

$BP=\sqrt{(2-x)^2+(-1-y)^2}$

$AP=3BP\\\sqrt{(-1-x)^2+(1-y)^2}=3\sqrt{(2-x)^2+(-1-y)^2}$

On squaring both sides, we get

$(-1-x)^2+(1-y)^2=9\left [ (2-x)^2+(-1-y)^2 \right ]\\1+x^2+2x+1+y^2-2y=9\left ( 4 +x^2-4x+1+y^2+2y\right )\\1+x^2+2x+1+y^2-2y=36+9x^2-36x+9+9y^2+18y\\8x^2+8y^2+43-38x+20y=0$

So, equation of curve is $8x^2+8y^2+43-38x+20y=0$

5 0

3 years ago