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3 years ago

Find m2ABC (3x + 8) (5x - 40)

Mathematics

1 answer:

Virty [35]3 years ago

7 0

Answer:

Step-by-step explanation:

(3x + 8) =(5x - 40)

Remove the brackets and replace 3x with -40 and -40 with 3x.

40 + 8 =5x - 3x (Remember that when you replace, the signs change from negative to positive and vise versa)

Solve for x. x= -24

Replace x with its value.

3(-24)+ 8= -64

5(-24) - 40= -160

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A plumber charges a $70 flat rate for all home repairs. His hourly rate is $15 per half hour. How much would a 2 hour and 26 m

Gnesinka [82]

The plumber's 2 hour and 26 min plumbing bill cost is $73

What is the plumber's per minute rate?

The plumber's per minute rate is the per half-hour rate divided by 30 minutes since the plumber charges $15 for every 30-minute job

per-minute rate=$15/30

per-minute rate=$0.50

Now, let us convert 2hours and 26 minutes to minutes, in other words, every 1 hour is 60 minutes, the two hours equal 120 minutes plus the 26 minutes gives a total of 146 minutes

Total minutes=120 min+26 min

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2 years ago

The fourth grade student at Riverside school are going on a field trip . There are 68 student on each of the four buses. How man

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3 years ago

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Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +

Veronika [31]

Answer:

Verified

$y(x) = \frac{3Ln(x) + 3}{x}$

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}$

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

$x^2y' +xy = 3$

- A general solution to the above ODE is also given as:

$y = \frac{3Ln(x) + C }{x}$

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

$y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}$

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

$-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3$

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3$

- Therefore, the complete solution to the given ODE can be expressed as:

$y ( x ) = \frac{3Ln(x) + 3 }{x}$

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)$

- Therefore, the complete solution to the given ODE can be expressed as:

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}$

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3 years ago

What is 7x6 like 7 rows of 6 collums

Darina [25.2K]

I guess it's 42
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Please help me understand the question

Sholpan [36]

Answer:

○ $\displaystyle \textcolor{black}{C.}\:triangle$

Explanation:

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I am joyous to assist you at any time.

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2 years ago