Answer:
D. (1/4, -2)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
y = -8x
4x - y = 3
<u>Step 2: Solve for </u><em><u>x</u></em>
<em>Substitution</em>
- Substitute in <em>y</em>: 4x - (-8x) = 3
- Simplify: 4x + 8x = 3
- Combine like terms: 12x = 3
- Isolate <em>x</em>: x = 3/12
- Simplify: x = 1/4
<u>Step 3: Solve for </u><em><u>y</u></em>
- Define equation: y = -8x
- Substitute in <em>x</em>: y = -8(1/4)
- Multiply: y = -2
Answer:
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Add up all the sides and you get 10
Answer:
D.
Step-by-step explanation:
Remember that the limit definition of a derivative at a point is:
![\displaystyle{\frac{d}{dx}[f(a)]= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28a%29%5D%3D%20%5Clim_%7Bx%20%5Cto%20a%7D%5Cfrac%7Bf%28x%29-f%28a%29%7D%7Bx-a%7D%7D)
Hence, if we let f(x) be ln(x+1) and a be 1, this will yield:
![\displaystyle{\frac{d}{dx}[f(1)]= \lim_{x \to 1}\frac{\ln(x+1)-\ln(2)}{x-1}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%281%29%5D%3D%20%5Clim_%7Bx%20%5Cto%201%7D%5Cfrac%7B%5Cln%28x%2B1%29-%5Cln%282%29%7D%7Bx-1%7D%7D)
Hence, the limit is equivalent to the derivative of f(x) at x=1, or f’(1).
The answer will thus be D.
Answer: FALSE
<u>Step-by-step explanation:</u>
x⁵ + 0x⁴ - 4x³ + 2x² - 4x + 1
If x - 2 is a factor, then x =2 should result in a remainder of 0 when using synthetic division.
2 | 1 0 -4 2 -4 1
<u>| ↓ 2 4 0 4 0 </u>
1 2 0 2 0 1 ← Remainder ≠ 0 so x - 2 is NOT a factor
