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3 years ago

13

Solve for the indicated variable. include all of your work in your answer. submit your solution. p = 3 r + 2 s ; for s

2 answers:

Lelu [443]3 years ago

5 0

The answer to the question

saveliy_v [14]3 years ago

3 0

p = 3r + 2s

p - 3r = 2s <em>subtract 3r from both sides</em>

(p - 3r)/2 = s <em>divide both sides by 2</em>

Answer: (p - 3r)/2

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Read 2 more answers

Which is the maximum value of the function y=-3x2 - 15x+9?

blagie [28]

Answer:

B

Step-by-step explanation:

The maximum value of the function is the y- coordinate of the vertex.

Given a parabola in standard form

y = ax² + bx + c ( a ≠ 0 )

Then the x- coordinate of the vertex is

$x_{vertex}$ = - $\frac{b}{2a}$

y = - 3x² - 15x + 9 ← is in standard form

with a = - 3 and b = - 15, thus

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Substitute x = - 2.5 into the equation for y- coordinate of vertex

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3 years ago

When the nth root of a is written, it is the positive value that is shown

allsm [11]

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Find the indicated nth partial sum of the arithmetic sequence.<br> −7, −3, 1, 5, . . ., n = 30

TiliK225 [7]

Step-by-step explanation:

Given the Arithmetic sequence

$-7, -3, 1, 5, . . .$

An arithmetic sequence has a constant difference d and is defined by

$a_n=a_1+\left(n-1\right)d$

$\mathrm{Compute\:the\:differences\:of\:all\:the\:adjacent\:terms}:\quad \:d=a_{n+1}-a_n$

$-3-\left(-7\right)=4,\:\quad \:1-\left(-3\right)=4$

$\mathrm{The\:difference\:between\:all\:of\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}$

$d=4$

$\mathrm{The\:first\:element\:of\:the\:sequence\:is}$

$a_1=-7$

as

$a_n=a_1+\left(n-1\right)d$

$\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:$

$a_n=4\left(n-1\right)-7$ ∵ $d=4$

$\mathrm{Arithmetic\:sequence\:sum\:formula:}$

$n\left(a_1+\frac{d\left(n-1\right)}{2}\right)$

$\mathrm{Plug\:in\:the\:values:}$

$n=30,\:\space a_1=-7,\:\spaced=4$

$=30\left(-7+\frac{4\left(30-1\right)}{2}\right)$

$=30\left(58-7\right)$ ∵ $\frac{4\left(30-1\right)}{2}=58$

$=30\cdot \:51$

$=1530$ ∵ $\mathrm{Multiply\:the\:numbers:}\:30\cdot \:51=1530$

Therefore, the indicated nth partial sum of the arithmetic sequence is 1530.

ANOTHER METHOD

as

$a_n=4\left(n-1\right)-7$

n = 30

$\sum _{n=1}^{30}\:4\left(n-1\right)-7$

$=\sum _{n=1}^{30}4n-11$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \sum a_n+b_n=\sum a_n+\sum b_n$

$=\sum _{n=1}^{30}4n-\sum _{n=1}^{30}11$

as

$\sum _{n=1}^{30}4n=1860$

and

$\sum _{n=1}^{30}11=330$

so

$=1860-330$

$=1530$

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