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Solve for x:
3(x - 5) = 6
Distribute the 3 to the variables inside the parenthesis
3x - 15 = 6
Add 15 to both sides to cancel out the "-15"
3x = 21
Divide both sides by 3
x = 7
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Answer:
{x ∈ R: x<6}
Step-by-step explanation:
Given
- x is a real number
- x is less than 6
Required
Write the set using set builder notation
The very first thing to do is to list out the range of x, using inequalities;
x is less than 6 implies that -infiniti < x < 6
The next step is to translate this to set builder. This is done as follows
x ∈ R - > This means that x is a real number
x < 6 -> where x is less than 6.
Bringing these two together, it gives:
{x ∈ R: x<6}
Hence, the set of real numbers x less than 6 is equivalent to {x ∈ R: x<6} using set builder notation
Answer:
26.6°
Step-by-step explanation:
Answer:
Σ(-1)^kx^k for k = 0 to n
Step-by-step explanation:
The nth Maclaurin polynomials for f to be
Pn(x) = f(0) + f'(0)x + f''(0)x²/2! + f"'(0)x³/3! +. ......
The given function is.
f(x) = 1/(1+x)
Differentiate four times with respect to x
f(x) = 1/(1+x)
f'(x) = -1/(1+x)²
f''(x) = 2/(1+x)³
f'''(x) = -6/(1+x)⁴
f''''(x) = 24/(1+x)^5
To calculate with a coefficient of 1
f(0) = 1
f'(0) = -1
f''(0) = 2
f'''(0) = -6
f''''(0) = 24
Findinf Pn(x) for n = 0 to 4.
Po(x) = 1
P1(x) = 1 - x
P2(x) = 1 - x + x²
P3(x) = 1 - x+ x² - x³
P4(x) = 1 - x+ x² - x³+ x⁴
Hence, the nth Maclaurin polynomials is
1 - x+ x² - x³+ x⁴ +.......+(-1)^nx^n
= Σ(-1)^kx^k for k = 0 to n
