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dedylja [7]
3 years ago
12

Joseph and his younger brother, Ryan, are taking a backpacking trip and must take a total of 12 gallons of water. Ryan will carr

y 4 less than half of gallons Joseph will carry. Let x be the number of gallons Joseph carries and let y be the number of gallons Ryan carries.
Mathematics
1 answer:
tangare [24]3 years ago
8 0

Hope it helps you...

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David goes into a candy store with $5.00. he buys 9 peppermints for $0.15 each, and some sour candies. each sour candy costs $0.
ANEK [815]
Step one. $5.00 - 9 peppermints
= $5.00 - $1.35
= $3.65
Step two. Figure out how many 25 cents are in $3.65.
$3.65 divided by .25
= 14.6 So David can buy 14 sour candies.
7 0
3 years ago
Explain your reasoning
exis [7]

Answer:

2 option= 3+3+3+3= 3(4) because it do not have exponent!!

Step-by-step explanation:

5 0
3 years ago
To which subset of real numbers does the number one-fourth belong? (1 point)
Trava [24]
It belongs to Q, The st of RATIONAL NUMBERS. It's written in terms of set:

1/4 ∈ Q={1/4, I 1∈Z, q∈Z, with q≠0}
8 0
4 years ago
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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6 0
3 years ago
What are the solutions of 12 - x2 =0?
butalik [34]
I hope that i helped ;D

3 0
2 years ago
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