Factoring: All Techniques CombinedFactor Each

4) $x^(6) + 2 x^(4) - 16 x^(2) - 32$

Answers

Answer 1

Answer: $x^(6) + 2 x^(4) - 16 x^(2) - 32 =x^4(x^2+2)-16(x^2+2)=(x^2+2)(x^4-16)=\n \n=(x^2+2)(x^4-4^2)=(x^2+2)(x^2-4)(x^2+4) =\n \n=(x^2+2)(x^2+4)(x+2)(x-2) \n \n \na^2-b^2 =(a-b)(a+b)$

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. Use the process outlined in the lesson to approximate the number 3√10. Use the approximation √10 ≈3162 277 7.. Find a sequence of five intervals that contain 3√10 whose endpoints get successively closer to 3√10. Write your
iintervals in the form 3^???? < 3^√10 < 3^s for rational numbers ????and s.

Answers

Answer:

sequence of five intervals

(1) 3³ < $3^(√(10) )$ < $3^(4)$

(2) $3^(3.1)$ < $3^(√(10) )$ < $3^(3.2)$

(3) $3^(3.16)$ < $3^(√(10) )$ < $3^(3.17)$

(4) $3^(3.162)$ < $3^(√(10) )$ < $3^(3.163)$

(5) $3^(3.1622)$ < $3^(√(10) )$ < $3^(3.1623)$

Step-by-step explanation:

as per question given data

√10 ≈ 3.162 277 7

to find out

sequence of five intervals

solution

as we have given that √10 value that is here

√10 ≈ 3.162 277 7 ........................1

when we find $3^(√(10) )$ ................2

put here √10 value in equation number 2

we get $3^(√(10) )$ that is 32.27

sequence of five intervals

(1) 3³ < $3^(√(10) )$ < $3^(4)$

(2) $3^(3.1)$ < $3^(√(10) )$ < $3^(3.2)$

(3) $3^(3.16)$ < $3^(√(10) )$ < $3^(3.17)$

(4) $3^(3.162)$ < $3^(√(10) )$ < $3^(3.163)$

(5) $3^(3.1622)$ < $3^(√(10) )$ < $3^(3.1623)$

Susu is solving the quadratic equation 4x2 – 8x – 13 = 0 by completing the square. Her first four steps are shown in the table.In which step did Susu first make an error?

Step 1
Step 2
Step 3
Step 4

Answers

In the question "Susu is solving the quadratic equation 4x2 – 8x – 13 = 0 by completing the square. Her first four steps are shown in the table. In which step did Susu first make an error?" Susu made her first error in step 2, when she factored out 4 she is supposed to divide 8x by 4. Thus the correct expression for step 2 should be 4(x^2 - 2x) = 13 To complete the solution for the equation: 4(x^2 - 2x + 1) = 13 + 4(1) 4(x - 1)^2 = 13 + 4 4(x - 1)^2 = 17 (x - 1)^2 = 17/4 x - 1 = +/- sqrt(17/4) x = 1 +/- sqrt(17) / 2

Answer:B

Step-by-step explanation:

If f(x) = 3x + 10 and g(x) = 2x - 4, find (f+g)(x).O A. (f+g)(x) = -34 - 2x - 14
B. (f+ g)(x) = 3x - 2x + 14
O C. (f+ g)(x) = 5x + 6
D. (f+ g)(x) = 3x + 2x + 6

Help! I’m getting really confused which one it is

Answers

Answer:

O D

Step-by-step explanation:

f+g(x)

3x+10+2x-4

3x +2x+6

Solve each of the following equations for x.a) 3x - 8 =29 b) 3 ( x - 8 ) = 28

c) 3 (x - 8) + 17 =29 d) 7x + 12 = 3x - 8

Answers

$3x - 8 =29 \n \n 3x = 29 + 8 \ / \ add \ 8 \ to \ each \ side \n \n 3x = 37 \ / \ simplify \n \n x = (37)/(3) \ / \ divide \ each \ side \ by \ 3 \n \n Answer: \fbox {x = 37/3} \ or \ \fbox {x = 12.3333}$

--------

$3 ( x - 8 ) = 28 \n \n x - 8 = (28)/(3) \ / \ divide\ each \ side \ by \ 3 \n \n x = (28)/(3) + 8 \ / \ add \ 8 \ to \ each \ side \n \n x = (52)/(3) \ / \ simplify \n \n Answer: \fbox {x = 52/3} \ or \ \fbox {x = 17.3333}$

--------

$3 (x - 8) + 17 =29 \n \n 3(x - 8) = 29 - 17 \ / \ subtract \ 17\ from \ each \ side \n \n 3(x - 8) = 12\ / \ simplify \n \n x - 8 = (12)/(3) \ / \ divide \ each \ side \ by \ 3 \n \n x - 8 = 4 \ / \ simplify \n \n x = 4 + 8 \ / \ add \ 8 \ to \ each \ side \n \n x = 12 \ / \ simplify \n \n Answer: \fbox {x = 12}$

--------

$7x + 12 = 3x - 8 \n \n 7x + 12 - 3x = -8 \ / \ subtract \ 3x \ from \ each \ side \n \n 4x + 12 = -8 \ / \ simplify \n \n 4x = -8 - 12 \ / \ subtract \ 12 \ from \ each \ side \n \n 4x = -20 \ / \ simplify \n \n x = - (20)/(4) \ / \ divide \ each \ side \ by \ 4 \n \n x = -5 \ / \ simplify \n \n Answer: \fbox {x = -5}$

The top and bottom margins of a poster are each 15 cm and the side margins are each 10 cm. If the area of printed material on the poster is fixed at 2400 cm2, find the dimensions of the poster with the smallest area.--------- cm (width)--------- cm (height)

Answers

Answer:

Step-by-step explanation:

let the sides be x and y

x y=2400

$y=(2400)/(x)\nnew~ dimensions ~are ~x+30~and~y+20\narea~A=(x+30)(y+20)\n=xy+20x+30y+600\n=2400+600+20x+30*(2400)/(x)\n=3000+20x+(72000)/(x)\n(dA)/(dx)=20-(72000)/(x^2)\n(dA)/(dx)=0~gives\n20 x^2=72000\nx^2=3600\nx=√(3600) =60\n(d^2A)/(dx^2)=(144000)/(x^3)>0~at ~x=60\ny=(2400)/(60)=40\n$

so A is minimum when x=60

y=40 cm

so dimensions are 60+30=90 cm

and 40+20=60 cm

Final answer:

The dimensions of the poster that provide the smallest total area, while maintaining a fixed printed area of 2400 cm², are 80 cm in width and 70 cm in height. This is obtained by applying calculus to optimize the area function of the poster.

Explanation:

The subject of this question is related to optimizing the area of a rectangular poster by adjusting its dimensions. Given that the area of printed material is fixed at 2400 cm², let's denote the width of the printed area as x (in cm) and so its height will be 2400/x (in cm).

Therefore, the total area of the poster, including margins, would be $(x+20)(2400/x + 30).$ We want to minimize this area. This is a calculus problem - take the derivative of the area with respect to x, set it equal to zero and solve for x. You'll obtain two possible dimensions for the width of the printed area: 40 cm and 60 cm. By testing these in the second derivative, you'll find that a width of 60 cm gives the minimum area. Therefore, the dimensions of the poster that gives the smallest total area are 60+20=80 cm (width) and $2400/60+30=70$ cm (height).

Learn more about Optimizing Area here:

brainly.com/question/30383510

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I need help to graphs these problem please

Answers

If you're talking about #5, where you have to graph the equation . . .

All you have to do is find two points on the graph, and draw a line between them.

How do you find points ? That's where you are in control !
Just pick any number you like for 'x', and then use the equation
to calculate the 'y' that goes with it.

Examples:

Pick 'zero' for 'x'.
Y = (1/2)x + 3
Y = 3

Pick 10 for 'x'.
y = (1/2)x + 3
y = 5 + 3
y = 8

Just create two points that way, using the equation. Then mark
the two points on the graph, and draw a line through the two points.

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