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strojnjashka [21]
4 years ago
14

If u(x) = x^5 - x^4 + x^2 and v (x) = -x^2 which expression is equivalent to ( u/ v) (x)

Mathematics
2 answers:
yawa3891 [41]4 years ago
3 0

Answer:

(u/v)(x)=-x^{3}+x^{2}-1

Step-by-step explanation:

You have the following functions:

u(x)=x^{5}-x^{4}+x^{2}\\v(x)=-x^{2}

 Therefore (u/v)(x) indicates that you must divide both functions, as you can see below:

(u/v)(x)=\frac{x^{5}-x^{4}+x^{2}}{-x^{2}}

Simplify it. Therefore, you obtain:

(u/v)(x)=\frac{x^{5}-x^{4}+x^{2}}{-x^{2}}\\\\(u/v)(x)=-x^{3}+x^{2}-1

balandron [24]4 years ago
3 0

Answer:

(u/v)(x) = -x³+x²-1

Step-by-step explanation:

We have given two functions.

u(x) = x⁵ - x⁴ + x² and v (x) = -x²

We have to find the quotient of the given two functions.

(u/v)(x) = ?

The formula to find the quotient is:

(u/v)(x) = u(x) / v(x)

Putting given values in above formula, we have

(u/v)(x) = x⁵-x⁴+x² / -x²

(u/v)(x) = -x²(-x³+x²-1) / -x²

(u/v)(x) = -x³+x²-1 which is the answer.

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Are the equations 3x= -9 and 4x= -12 equivalent
Hitman42 [59]
Yes they are equivalent. You get -3 when you try solving the actual equation.
5 0
4 years ago
Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|
Fynjy0 [20]

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

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Simplify the algebraic expression (4b5 - 7) + (12b5 - 1)
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Step-by-step explanation:

4 0
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