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2 years ago

G(x)=x3+6x2+12x+8 when x=-1

1 answer:

3 0

G=38
Whenever you divide or multiply by one the number stays the same, therefore just take out all the xs and then solve the equation. I hope this helps!

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What is the answer for 6?

marusya05 [52]

IS the answer ii and iv

and the other is y<5.3

7 0

2 years ago

Please help!!!!!!!!!!

steposvetlana [31]

9514 1404 393

Answer:

-203.1875

Step-by-step explanation:

The ends of the interval are ...

(-4, g(-4)) = (-4, 1023 15/16)

(1, g(1)) = (1, 8)

The average rate of change is the slope of the line between these two points:

m = (y2 -y1)/(x2 -x1)

m = (8 -(1023 15/16))/(1 -(-4)) = (-1015 15/16)/5 = -203 3/16

The average rate of change on the interval is -203 3/16.

5 0

2 years ago

Jason can make 7 origami cranes in 10 minutes. How many minutes will it take to make 1505 origami cranes

Angelina_Jolie [31]

Answer:

The unit rate = 7/10 cranes per minute

Step-by-step explanation:

So, the number of minutes would it would take Kenny to fold 1505 cranes is

10/7 x 1505 = 2150 minutes

2150/60 = 35 5/6 hours

8 0

1 year ago

Read 2 more answers

Help appreciated on question in image!<br> Thanks:)

Verdich [7]

Answer:

$x=-1,\:x=-7,\:x=i,\:x=-i$

Step-by-step explanation:

Considering the equation

$x^4+8x^3+8x^2+8x+7=0$

Solving

$x^4+8x^3+8x^2+8x+7$

$\mathrm{Factor\:}x^4+8x^3+8x^2+8x+7:\quad \left(x+1\right)\left(x+7\right)\left(x^2+1\right)$

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=7,\:\quad a_n=1$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:7,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:7}{1}$

$-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1$

$=\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]$

Solving

$\frac{x^4+8x^3+8x^2+8x+7}{x+1}$

$=x^3+7x^2+x+7$

Putting $\frac{x^4+8x^3+8x^2+8x+7}{x+1}$ = $x^3+7x^2+x+7$ in equation [A]

So,

$\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]$

$=\left(x+1\right)x^3+7x^2+x+7$

$x^3+7x^2+x+7=\left(x+7\right)\left(x^2+1\right)$

So,

Equation [A] becomes

$=\left(x+1\right)\left(x+7\right)\left(x^2+1\right)$

So, the polynomial equation becomes

$\left(x+1\right)\left(x+7\right)\left(x^2+1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$ $\mathrm{Solve\:}\:x+1=0:\quad x=-1$