Draw a model to help solve 5/6 + 1/4. Write your anserw as a mixed number

What is the product of 2X plus 3 and 4X squared minus 5X plus 6

Tju [1.3M]

The product of something means multiplying the terms together.

(2x+3) (4x^2-5x+6)

Secondly you need to distribute the terms to each other (Think of problems like FOIL)

2x * 4x^2 + 2x(-5x) + 2x * 6 + 3 * 4x^2 + 3(-5x) + 3 * 6

Then you must take into account that some of the numbers are negative. (minus-plus rules!)

2x * 4x^2 - 2x * 5x + 2x * 6 + 3 * 4x^2 - 3 * 5x + 3 * 6

Now is the tricky part of simplifying everything.

2x * 4x^2 = 8x^3

2x * 5x = 10x^2

2x * 6 = 12x

3 * 4x^2 = 12x^2

3 * 5x = 15x

3 * 6 = 18

8x^3 - 10x^2 + 12x + 12x^2 - 15x + 18

Then you group like terms.

8x^3 - 10x^2 + 12x^2 - 3x + 18

8x^2 + 2x^2 - 3x + 18

The trickiest part of this is distributing all of the terms within the parentheses, once you've done that, it's smooth sailing!

5 0

3 years ago

1. A sine function has the following key features:

andrew11 [14]

Problem 1

See the attached image (figure 1)

16pi seems like a typo. I'm going to assume that it's a fraction and it is 1/(6pi)
f = 1/(6pi) = frequency
T = 1/f = 1/(1/(6pi)) = 6pi
Amplitude = 2
a = 2
b = 2pi/T = 2pi/(6pi) = 1/3
Midline: y = 3
d = 3

The function is
y = a*sin(bx-c)+d
y = 2*sin(1/3*x-0)+3
y = 2*sin(x/3)+3

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Problem 2

See the attached image (figure 2)

T = 12 is the period
a = 4 is the amplitude
b = 2pi/T = 2pi/12 = pi/6
y = 1 is the midline so d = 1
The y intercept is (0,1) which is the midline, which indicates no phase shifts have occurred so c = 0

The function is
y = a*sin(bx-c)+d
y = 4*sin((pi/6)x-0)+1
y = 4*sin((pi/6)x)+1

===============================================

Problem 3

See the attached image (figure 3)

Period = 4pi
T = 4pi
b = 2pi/T = 2pi/(4pi) = 1/2 = 0.5
Amplitude = 2
a = 2
Midline: y = 3
d = 3
y-intercept: (0,3)
The function is a reflection of its parent function over the x-axis, so 'a' is negative meaning a = -2 instead of a = 2

The function is
y = a*sin(bx-c)+d
y = -2*sin(0.5x-0)+3
y = -2*sin(0.5x)+3

===============================================

Problem 4

See the attached image (figure 4)

a = 10 which is half of the distance between the highest and lowest points
T = 8 is the period
b = 2pi/T = 2pi/8 = pi/4
c = -pi/2 is the phase shift since its really a cosine graph
d = 0 is the midline

The function is
y = a*sin(bx-c)+d
y = 10*sin((pi/4)*x+(-pi/2))+0
y = 10*sin((pi/4)*x+pi/2)

===============================================

Problem 5

See the attached image (figure 5)

a = 2 is the amplitude since it bobs up and down this distance from the midline
T = 8 seconds is the period (double that of the time it takes for it to go from the highest to the lowest point)
b = 2pi/T = 2pi/8 = pi/4
c = 0 is the phase shift as the buoy starts at normal depth of 20 meters
d = 20 is the midline

The function is
y = a*sin(bx-c)+d
y = 2*sin((pi/4)x-0)+20
y = 2*sin((pi/4)x)+20

===============================================