Answer:
2 and -3
Step-by-step explanation:
2x^2=4x
4x+2x=12
6x=12
how many times does 6 go into 12?
right, 2 times!
x=2
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2x^2+2x-12=0
x^2+x-6=0
(x+3)(x-2)=0
X=-3 and 2
Answer:
it should be -4+y
Step-by-step explanation:
Answer is B. CSC A= 13 over 6
You need to learn these basic rules:
Sin is opposite csc
Cos is opposite sec
Tan is opposite Cot
In saying this, CsC is the reciprocal for Sin. You would have to switch the numerator and denominatior to get your answer
Answer:
11.2 feet
Step-by-step explanation:
The two furthest corners of the bed are at (8, 6) and (5, 10). To determine which is the furthest from the origin, use the distance formula.
d = √((x₂ − x₁)² + (y₂ − y₁)²)
d = √((8 − 0)² + (6 − 0)²)
d = 10
d = √((x₂ − x₁)² + (y₂ − y₁)²)
d = √((5 − 0)² + (10 − 0)²)
d = 5√5
d ≈ 11.2
Therefore, the corner at (5, 10) is the furthest from the origin, about 11.2 feet away.
Answer:
R3 <= 0.083
Step-by-step explanation:
f(x)=xlnx,
The derivatives are as follows:
f'(x)=1+lnx,
f"(x)=1/x,
f"'(x)=-1/x²
f^(4)(x)=2/x³
Simialrly;
f(1) = 0,
f'(1) = 1,
f"(1) = 1,
f"'(1) = -1,
f^(4)(1) = 2
As such;
T1 = f(1) + f'(1)(x-1)
T1 = 0+1(x-1)
T1 = x - 1
T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2
T2 = 0+1(x-1)+1(x-1)^2
T2 = x-1+(x²-2x+1)/2
T2 = x²/2 - 1/2
T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3
T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3
T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)
Thus, T1(2) = 2 - 1
T1(2) = 1
T2 (2) = 2²/2 - 1/2
T2 (2) = 3/2
T2 (2) = 1.5
T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)
T3(2) = 4/3
T3(2) = 1.333
Since;
f(2) = 2 × ln(2)
f(2) = 2×0.693147 =
f(2) = 1.386294
Since;
f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).
Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,
Since;
f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2
Finally;
R3 <= |2/(4!)(2-1)^4|
R3 <= | 2 / 24× 1 |
R3 <= 1/12
R3 <= 0.083
