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4 years ago

Does 4 13/20 terminate or repeat

Mathematics

1 answer:

miskamm [114]4 years ago

8 0

4 13/20 terminates. When you divide 13 by 20 you get 0.65 which terminates.

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Is the square root of 42.25

liraira [26]

Answer: It’s 6.5

Step-by-step explanation: Bruh just look it up on google

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3 years ago

Identify each matrix A such that A^2 has identical diagonal elements?

Maksim231197 [3]

Answer:

The correct matrices are:

Matrix:

7 1 5

1 5 7

5 7 1

all diagonal elements of A^2 are: 7^2 + 1^2 + 5^2

Matrix:

9 18 27

27 -9 18

18 27 9

all diagonal elements of A^2 are: 9^2 + 27*18 + 18*27 or (-9)^2

Matrix:

8 1 6

6 8 1

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all diagonal elements of A^2 are: 8^2 + 6*1 + 1*6

6 0

4 years ago

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Gennadij [26K]

Answer:The areas is 102m

Step-by-step explanation:

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3 years ago

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Which transformation of triangle T will produce triangle U?

Bezzdna [24]

Answer:A

Step-by-step explanation:

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3 years ago

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find

9966 [12]

Answer:

The particular integral of given differential equation

$y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))$

General solution of given differential equation

$y = y_{c} + y_{p}$

$Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))$

Step-by-step explanation:

Step(i):-

Given Differential equation y'' − 5 y' + 4 y = x

Given equation in operator form

D²y - 5 Dy + 4 y = x

⇒ ( D² - 5 D + 4 ) y =x

⇒ f(D) y = Q

where f(D) = D² - 5 D + 4 and Q(x) = x

The auxiliary equation f(m) =0

m²-5 m + 4 =0

m² - 4 m - m + 4 =0

m ( m -4 ) -1 ( m-4) =0

(m - 1) =0 and ( m-4) =0

m = 1 and m =4

The complementary function

$Y_{c} = C_{1} e^{x} + C_{2} e^{4x}$

Step(ii):-

particular integral

Particular integral

$y_{p} = \frac{1}{f(D)} Q(x) = \frac{1}{D^{2} - 5 D + 4} X$

taking common '4'

$= \frac{1}{4(1 + (\frac{D^{2} - 5 D}{4} ))} X$

$=\frac{1}{4} (1 + (\frac{D^{2} -5D}{4})^{-1} )} X$

applying binomial expression

( 1 + x )⁻¹ = 1 - x + x² - x³ +.....

$=\frac{1}{4} (1 - (\frac{D^{2} -5D}{4}) +((\frac{D^{2} -5D}{4})^{2} -...} )X$

Now simplifying and we will use notation D = $\frac{dy}{dx}$

$=\frac{1}{4} (x - (\frac{D^{2} -5D}{4})x +((\frac{D^{2} -5D}{4})^{2}(x) -...}$

Higher degree terms are neglected

$=\frac{1}{4} (x - (\frac{ -5 D}{4}) x)$

The particular integral of given differential equation

$y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))$

Final answer:-

General solution of given differential equation

$y = y_{c} + y_{p}$

$Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))$

4 0

3 years ago