We use the distance formula for this problem.
d = √[(x₂ - x₁)² + (y₂ - y₁)²]
The distance between point (-2,-2) and point (-2,4).
d = √[(⁻2 - ⁻2)² + (4 - ⁻2)²] = 6 units
Then, compute for 20% of 6 units:
Distance traveled = 6(0.2) = 1.2 units
Use 1.2 units as distance and the starting point (-2,-2). The x-coordinate should still be at -2 because the distance is a straight line as shown in the picture.
1.2 = √[(-2 - ⁻2)² + (y - ⁻2)²]
Solving for y,
y = -0.8
The point is found at (-2,-0.8). This is located at quadrant 3. As to the distance traveled, that would be: 1.2*6 = 6 miles. Thus, the answer is C.
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
Answer:
20
Step-by-step explanation:
an integer m is divisible by both 10 and 12
Lets write the prime factor for each number
the prime factor for 10 is 5,2
the prime factor for 12 is 3,2,2
Now take the product of the prime factors that are not in common and also in common
60 is divisible by 20
so answer is 20
