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3 years ago

What is the ninth term in A(n)=-4 - 6(n-1)?

Mathematics

1 answer:

Ghella [55]3 years ago

8 0

Answer:

A(9) = -52

Step-by-step explanation:

The given sequence is :

A(n)=-4 -6(n-1)

We need to find the 9th term of the above sequence.

To find it, put n = 9 in the above sequence.

A(9) = -4 -6(9-1)

= -4-6(8)

= -4-48

= -52

Hence, the 9th term of the sequence is -52.

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Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (0, 11, −8) and

NNADVOKAT [17]

Answer:

The vector equation of the line is $\overrightarrow {r}=(11j-8k)+t(4i-4j+6k)$ and parametric equations for the line are $x=4t$ , $y=11-4t$ , $z=-8+6t$ .

Step-by-step explanation:

It is given that the line passes through the point (0,11,-8) and parallel to the line

$x=-1+4t$

$y=6-4t$

$z=3+6t$

The parametric equation are defined as

$x=x_1+at,y=y_1+bt,z=z_1+ct$

Where, (x₁,y₁,z₁) is point from which line passes through and <a,b,c> is cosine of parallel vector.

From the given parametric equation it is clear that the line passes through the point (-1,6,3) and parallel vector is <4,-4,6>.

The required line is passes through the point (0,11,-8) and parallel vector is <4,-4,6>. So, the parametric equations for the line are

$x=4t$

$y=11-4t$

$z=-8+6t$

Vector equation of a line is

$\overrightarrow {r}=\overrightarrow {r_0}+t\overrightarrow {v}$

where, $\overrightarrow {r_0}$ is a position vector and $\overrightarrow {v}$ is cosine of parallel vector.

$\overrightarrow {r}=(11j-8k)+t(4i-4j+6k)$

Therefore the vector equation of the line is $\overrightarrow {r}=(11j-8k)+t(4i-4j+6k)$ and parametric equations for the line are $x=4t$ , $y=11-4t$ , $z=-8+6t$ .

3 0

3 years ago

Find two numbers which have a sum of 132 and that can be written as the simplified<br> ratio 3:8

Molodets [167]

Answer:

96 and 36

Step-by-step explanation:

3+8 = 11

132/11 = 12

so:

12 x 3 = 36

12 x 8 = 96