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Rzqust [24]
3 years ago
11

Given that R= 6x + 5y

Mathematics
1 answer:
Hunter-Best [27]3 years ago
7 0

Answer:

-28/5=y

Step-by-step explanation:

20=6(8)+5y

20=48+5y

-28=5y

-28/5=y

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F(x) = 3^x; find f(0)​
spin [16.1K]

Answer:

f(0)=1

Step-by-step explanation:

The question asks us to find f(0). We want to find what f(x), or y is, when x is equal to 0.

f(x)=3^x

Therefore, we can substitute 0 in for x.

f(0)=3^0

Evaluate the exponent, 3^0.

Any number raised to the power of zero is equal to 1.

f(0)=1

7 0
3 years ago
Read 2 more answers
How to find y coordinate? Here are my x coordinates and this is the given formula:
Luda [366]
Y=x-3x+1
y=2-3(-4)+1
y=15
6 0
3 years ago
|y| x-3 <br> what is the vertex
lutik1710 [3]

Answer:

(3,0)

Step-by-step explanation:

put x-3=0 and solve for it whic hwould be x=3

8 0
3 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}

                           

Download docx
6 0
3 years ago
Terry Barlett worked from 7:00 am to 10:45 am and from 12:30 pm to 3:15 pm at a chocolate shop. Find the total hours
irakobra [83]

Answer:

6 hours 30 minutes

Step-by-step explanation:

morning work=3 hours 45 minutes

afternoon work=2 hours 45 minutes

total hours=?

Here,

3 hours 45 minutes+2 hours 45 minutes

=6 hours 30 minutes

Therefore,Terry Baelett worked 6hours 30 minutes a day.

6 0
2 years ago
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