Use place value to find 748-319

Setler [38]

If the question had asked what is 9/9 closer to you would say 1. When you increase the numerator by one digit go higher than 1.
10/9 = 1.11111111

so it is closer to 1 than to anything else

6 0

3 years ago

Two cubes have surface areas of 81 square inches and 144 square inches. What is the ratio of the volume of the small cube to the

Katena32 [7]

We know, surface area of cube = 6a²

So, side length of 1st cube = a² = 81/6
a = 3.67

Side length of 2nd cube = a² = 144/6
a = 4.90

Now, volume of cube = a³

Volume of 1st cube = 3.67³ = 49.43

Volume of 2nd cube = 4.90³ = 117.64

So, Ratio = 49.43 / 117.64 = 1/2.38

In short, Your Answer would be: 1: 2.38

Hope this helps!

5 0

3 years ago

Can someone help me with this

krok68 [10]

3 + 6 ( 4 + 3 ) is the answer

8 0

3 years ago

Read 2 more answers

The cost of a taxi ride is a linear function of the distance traveled. If a 5-mile ride costs $12 and a 9-mile ride costs $14, w

dalvyx [7]

it should be c=0.5m+9.5

8 0

3 years ago

Read 2 more answers

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