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WITCHER [35]
3 years ago
8

How do I add exponent denominator and numerator

Mathematics
2 answers:
earnstyle [38]3 years ago
4 0

Answer:If terms have the same base a and same fractional exponent n/m, we can add them. The rule is given as:

Can/m + Dan/m = (C + D)an/m

Here’s an example of adding fractional exponents:

2x2/5 + 7x2/5 = 9x2/5

Step-by-step explanation: anyway, i hope that this answers your question. Have a good day!

Leokris [45]3 years ago
3 0

Answer:

multiply

Step-by-step explanation:

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if one scale which means length of one small square is 1/4. we can see that the base of A has 4 scales with a value of 1/4. so to get the length you multiply the scale which is number of scales in the base which 4 if you look at the base. 1/4 x 4 is 1. to find height you multiply the scale with the number of scales upward which you can see has 3 scale so height is 1/4 x 3 = 3/4. you apply the same method to the other two to get the answer.

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Use the annihilator method to determine the form of a particular solution for the given equation. u double prime minus 2 u prime
sladkih [1.3K]

Answer:

the particular solution is

Y_{p}= C +D\sin 5t +E\cos 5t + F\exp 4t + G\exp -2t

the differential operator that annihilate the non homogeneous differential equation is

D(D^2+5)

Step-by-step explanation:

hello,

i believe the non homogeneous differential equation is

U^{''}  - 2U^{'} - 8= \cos 5x + 7

the homogeneous differential equation of the above is

u^{''} -2u^{'} -8 =0

the differential form of the above equation is

D^2-2D-8=0

(D-4)(D+2)=0

thus the roots are 4 and -2.

thus the solution of the homogenous differential equation is given as

Y_{h} (t)= A\exp{4t} + B\exp{-2t}

the differential operator of the non homogeneous equation is given as

(D-4)(D+2)(u)=\cos 5x +7

the differential operator D^2 +5 annihilates \cos 5x and the differential operator D annihilates 7

applying D(D^2+5) to both sides of the differential equation we have;

(D-4)(D+2)(u)=\cos 5x +7

D(D^2+5)(D-4)(D+2)=D(D^2+5)(\cos5x+7)D(D^2+5)(D-4)(D+2)=0

the roots of the characteristic polynomial of the diffrential equation above are 0, \cmplx 5i, -\cmplx 5i, 4, -2

thus the <em><u>particular solution</u></em> is

Y_{p}= C\exp{0}+D\sin 5t +E\cos 5t + F\exp {4t}  + G\exp {-2t}

this gives us the <em><u>particular solution</u></em>

Y_{p}= C +D\sin 5t +E\cos 5t + F\exp 4t + G\exp -2t

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3 years ago
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