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3 years ago

What's the rule about adding intergers?

Mathematics

1 answer:

vivado [14]3 years ago

5 0

Answer:

Integers include both positive and negative numbers, and there are several rules for adding integers. Adding two positive integers together will always result in a positive integer. For example, 6 + 4 = 10. To add two negative integers, add the numbers together and place a negative sign in front of the answer.

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Pooja's plant began sprouting 222 days before Pooja bought it, and she had it for 989898 days until it died. At its tallest, the

MakcuM [25]

Answer:

Step-by-step explanation:

Given:

function h(t) models the height of Pooja's plant (in cm) where

t is the number of days after she bought it.

Now we have to find about which number type is more appropriate for the domain of h.

That means what values can be taken by the variable "t".

As per the questions, t represents number of days not the hours so it will not be in decimal or fraction value. It can only use integer values for the number of days. like 1, 2, 3 ...,n

Now as the time is counted after she bought the plant then number of days will be positive.

Hence answer for the type of domain can be positive integers or you can say integers greater than or equal to 0.

5 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:

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