All categories

3 years ago

Write the explicit rule for the nth term. Then find a71. {O, -8, -16, -24 ...}

Mathematics

1 answer:

yulyashka [42]3 years ago

3 0

Answer:

Step-by-step explanation:

{0, -8, -16, -24,...} here you add -8 each time so the rule is

$a_{1} = 0, a_{n} = a_{n-1} -8$

the 71 term is

$a_{71} = 0 + (71 - 1) * (-8) = 70 * (-8) = -560$

You might be interested in

<img src="https://tex.z-dn.net/?f=%5Csqrt%7B18%7D%2B%5Csqrt%7B50%7D" id="TexFormula1" title="\sqrt{18}+\sqrt{50}" alt="\sqrt{18}

KengaRu [80]

$3\sqrt{2} +5\sqrt2}\\=8\sqrt{2}$

3 0

3 years ago

Read 2 more answers

I NEED HELP PLEASE, THANKS! :)

Alenkasestr [34]

Answer:

Option A

Step-by-step explanation:

This is a great question!

The first thing we want to do here is to graph the system of " inequalities " -

$\begin{bmatrix}2x+9y\le \:100\\ 9x+y\le \:54\end{bmatrix}$

As x is ≥ 0, respectively y ≥ 0, it makes things a lot simpler, as it kind of restricts us to quadrant 1, but not entirely.

First change each inequality to " slope - intercept " form, as such -

$2x + 10y \leq 100,\\10y \leq 100 - 2x,\\y \leq - 1 / 5x + 10\\----------\\9x + y \leq 54,\\y \leq - 9x + 54$

The graphed solutions would be in the attachment. I have colored the region with which the two intersect, and the fact that we are limited to quadrant 1. In this colored region there are 3 major points, ( 5, 9 ), ( 0, 10 ), and ( 6, 0 ). We have to determine which of these are are maximum points, as the minimum are the same. Substitute these values ( ( 5, 9 ), ( 0, 10 ), and ( 6, 0 ) ) into the equation f( x, y ) = 10x + 4y,

$f ( x, y ) = 10( 5 ) + 4( 9 ),\\f ( x, y ) = 86 -\\f ( x, y ) = 10( 0 ) + 4( 10 ),\\f ( x, y ) = 40 -\\f ( x, y ) = 10( 6 ) + 4( 0 ),\\f ( x, y ) = 60$

( 5, 9 ) resulted in the greatest amount, 86. That would make it our maximum point -

Solution = Option A

3 0

3 years ago

Examine the system of equations.

MrMuchimi

answer:

$x = - 2 \\ y = 2 \\ \\ ( - 2, \: 2)$

explanation:

$y = - \frac{1}{2} x + 1 \\ y = 2x + 6 \\ first \: equation \: add \: \frac{1}{2} from \: both \\ sides \\ \\ second \: equation \: subtract \: 2x \: from \\ both \: sides \\ \\ y + \frac{1}{2} x = 1 \\ y - 2x = 6 \\ solve \: the \: first \: equation \\ \\ y + \frac{1}{2} x = 1 \\ subtract \: \frac{x}{2} from \: both \: sides \\ \\ y = - \frac{1}{2} x + 1 \\ substitute \: - \frac{x}{2} + 1 \: for \: y \: in \: the \\ other \: equation \\ \\ - \frac{1}{2} x + 1 - 2x = 6 \\ add \: \frac{1}{2} x \: to \: 2x \\ \\ - \frac{5}{2} x + 1 = 6 \\ subtract \: 1 \: from \: both \: sides \\ \\ - \frac{5}{2} x = 5 \\ divide \: both \: sides \: by \: - \frac{5}{2} \\ \\ x = - 2 \\ substitute \: the \: value \: of \: x \: into \\ an \: equation \\ \\ y = - \frac{1}{2} ( - 2) + 1 \\ distribute \\ \\ y = 1 + 1 \\ add \\ \\ y = 2 \\ \\ ( - 2, \: 2)$

7 0

3 years ago

Find the second of two consecutive integers if the second is 13 less than twice the first.

emmasim [6.3K]

What is the range in which the integers can be in? I cannot fully answer this question. But I think there could be more than one answer. One answer would be the first is 36 and the second is 66.

Hope this helps!
Can u plz mark me as brainliest? I really need it!

4 0

3 years ago

Help Please. Can't seem to understand how to do this. I have tried and watch over 5 videos. Can't seem to get it.

Snezhnost [94]

Answer:

The solutions are the following:

z=2(cos(π6)+isin(π6))=√3+12i

z=2(cos(2π3)+isin(2π3))=−1+i√3

z=2(cos(7π6)+isin(7π6))=−√3−12i

z=2(cos(5π3)+isin(5π3))=1−i√3

hope this helps!! :) --Siveth

7 0

2 years ago