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3 years ago

3(4 - 2x) + 3x 12 + 3x 12 + 6x 12 - 3x 12 - 9x

1 answer:

3 0

3(4 - 2x) + 3x 12 + 3x 12 + 6x 12 - 3x 12 - 9x
12 - 6x + 3 • 12 + 3 • 12 + 6 • 12 - 3 • 12 - 9x
12 - 6x + 36 + 36 + 72 - 26 - 9x
-9x - 6x + 12 + 36 + 36 + 72 - 26
-15x + 156 - 26
-15x + 130

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Find the minimum or maximum of the function. Describe the domain/range in interval notation y=6x^2-1

Margarita [4]

max is at vertex

in form $y=ax^2+bx+c$

the x value of the vertex is $\frac{-b}{2a}$

given, $y=6x^2-1$

a=6, b=0

the x value of the vertex is -0/(2*6)=0

the y value is $y=6(0)^2-1=0-1=-1$

so vertex is at (0,-1)

since the value of a is positive, the parabola opens up and the vertex is a minimum value of the function

therefore that value is the smallest value the function can be

domain=numbesr you can use for x

range=numbesr you get out of inputting the domain

domain=all real numbers

range is {y | y≥-1} since y=-1 is the minimum

7 0

2 years ago

William got scores of q sub 1, , and q sub 3 on three quizzes.

USPshnik [31]

Answer:

a. The average is =(++) .

b. He can use the formula =−− q sub 3 , equals 3 x minus , q sub 1 , minus , q sub 2. He will need a score of =()−−= q sub 3 , equals 3 open 90 close minus 85 minus 88 equals 97 on his third quiz.

Step-by-step explanation:

3 0

3 years ago

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

2 years ago

Read 2 more answers

Find the area of the sector in<br> terms of pi.<br> 120°<br> 24<br> Area = [?] π<br> Enter

Nady [450]

Sector area= theta/360 x πr^2

Radius= 24/2= 12

Sector area= 120/360 x π(12)^2

Final answer :

Sector area = 48π

3 0

2 years ago

Sean used the $1,200 he got from his graduation party to open a savings account. If the account earns 1% interest each month and

Greeley [361]

Answer:

$\$1,920$

Step-by-step explanation:

we know that

The simple interest formula is equal to

$A=P(1+rt)$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

in this problem we have

$t=5*12=60\ months\\ P=\$1,200\\r=0.01$

substitute in the formula above

$A=\$1,200(1+0.01*60)$

$A=\$1,200(1.6)=\$1,920$

6 0

2 years ago