Answer:
a+2b-d=1, 3, 5, 7
Step-by-step explanation:
(ax^2+bx+3)(x+d)
ax^3+bx^2+3x+adx^2+bdx+3d
ax^3+bx^2+adx^2+3x+bdx+3d=x^3+6x^2+11x+12
ax^3=x^3, a=1
bx^2+adx^2=6x^2
x^2(b+ad)=6x^2
b+ad=6
b+(1)d=6
b+d=6
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3x+bdx=11x
x(3+bd)=11x
3+bd=11
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b=6-d
3+(6-d)d=11
3+6d-d^2=11
3-11+6d-d^2=0
-8+6d-d^2=0
d^2-6d+8=0
factor out,
(d-4)(d-2)=0
zero property,
d-4=0, d-2=0
d=0+4=4,
d=0+2=2
b=6-4=2,
b=6-2=4.
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a+2b-d=1+2(2)-2=1+4-2=5-2=3
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a+2(4)-4=1+8-4=9-4=5
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a+2(2)-4=1+4-4=5-4=1
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a+2(4)-2=1+8-2=9-2=7
For this case the amplitude is given by:
A = (l-1l + 3) / 2
A = 4/2
A = 2
The middle line is given by:
y = 1
The period of the function is given by:
T = l x2 - x1 l
T = l 0 - 2pi l
T = l - 2pi l
T = 2pi
Answer:
Amplitude: 2; period: 2π; midline: y = 1
Answer:
slope= 16
Step-by-step explanation:
use the slope formula= rise/run=. y2-y1/ x2-x1
Answer:
(-3, 4)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtract Property of Equality
<u>Algebra I</u>
- Terms/Coefficients
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
y = -x + 1
2x + 3y = 6
<u>Step 2: Solve for </u><em><u>x</u></em>
<em>Substitution</em>
- Substitute in <em>y</em>: 2x + 3(-x + 1) = 6
- Distribute 3: 2x - 3x + 3 = 6
- Combine like terms: -x + 3 = 6
- Isolate <em>x</em> terms: -x = 3
- Isolate <em>x</em>: x = -3
<u>Step 3: Solve for </u><em><u>y</u></em>
- Define equation: y = -x + 1
- Substitute in <em>x</em>: y = -(-3) + 1
- Simplify: y = 3 + 1
- Add: y = 4
Slope of line t is -1/5
If two lines are perpendicular, the product of their slopes is -1
