All categories

4 years ago

Can you add a negative and a positive when doing like term

Mathematics

2 answers:

Verdich [7]4 years ago

5 0

Yes, you can add a negative and a positive when doing like terms.

tester [92]4 years ago

3 0

In short yes. But to know for sure you can combine any term no matter the sign as long as they have the same type of variable attached to them for example

$x {}^{2} + x$
this cannot be combined because one is squared the other isn't
$x + - x$
you can combine these because they're the same type of variable this just becomes 1
$x + 3 + - 5$

we cant combine x into these because it's not just a constant number but we can combine 3 and negative 5 to get
$x - 2$

hope this helps!

You might be interested in

The graph of which of the following equations contains the points (2, 3) and (3, 2)?

IceJOKER [234]

Answer:

(2, 3) lies on the line y = 5 - x.

Step-by-step explanation:

Were there only two possible answer choices?

We can determine whether or not a given point, such as (2, 3), is contained in a function such as y = x - 1, by substituting the given coordinates 2 and 3 for x and y respectively; is the resulting equation true or false?

Check out (2, 3) and y = x - 1: 3 = 2 - 1, or 3 = 1. False. (2, 3) does not lie on the line y = x - 1. Does (2, 3) lie on the line y = 5 - x? Is 3 = 5 - 2 true? Yes. So (2, 3) lies on the line y = 5 - x.

Check out (2, 3) and y = 7 - 2x. Is 3 = 7 - 2(1) true? No, it is not true.

8 0

4 years ago

Are the ratios 8:16 and 2:4 equivalent? need answers quick!

drek231 [11]

Answer:

yes

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Between which two numbers is Negative 1.5 located on a number line?

Ede4ka [16]

Between -2&-1 is the answer

5 0

3 years ago

Read 2 more answers

❗️10 points❗️

astraxan [27]

Answer:

cubic trinomial

Step-by-step explanation:

Because the highest power of x is 3, this is a cubic trinomial.

6 0

3 years ago

In this problem you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+3y=8te^−t+6e^−t−(9t+6)

Luden [163]

We're given the ODE,

y'' + 4y' + 3y = 8t exp(-t ) + 6 exp(-t ) - (9t + 6)

(where I denote exp(x) = eˣ )

First determine the characteristic solution:

y'' + 4y' + 3y = 0

has characteristic equation

r ² + 4r + 3 = (r + 1) (r + 3) = 0

with roots at r = -1 and r = -3, so the characteristic solution is

y = C₁ exp(-t ) + C₂ exp(-3t )

For the non-homogeneous equation, assume two ansatz solutions

y₁ = (at ² + bt + c) exp(-t )

and

y₂ = at + b

• y'' + 4y' + 3y = 8t exp(-t ) + 6 exp(-t ) … … … [1]

Compute the derivatives of y₁ :

y₁ = (at ² + bt + c) exp(-t )

y₁' = (2at + b) exp(-t ) - (at ² + bt + c) exp(-t )

… = (-at ² + (2a - b) t + b - c) exp(-t )

y₁'' = (-2at + 2a - b) exp(-t ) - (-at ² + (2a - b) t + b - c) exp(-t )

… = (at ² + (b - 4a) t + 2a - 2b + c) exp(-t )

Substitute them into the ODE [1] to get

→ [(at ² + (b - 4a) t + 2a - 2b + c) + 4 (-at ² + (2a - b) t + b - c) + 3 (at ² + bt + c)] exp(-t ) = 8t exp(-t ) + 6 exp(-t )

(at ² + (b - 4a) t + 2a - 2b + c) + 4 (-at ² + (2a - b) t + b - c) + 3 (at ² + bt + c) = 8t + 6

4at + 2a + 2b = 8t + 6

→ 4a = 8 and 2a + 2b = 6

→ a = 2 and b = 1

→ y₁ = (2t ² + t ) exp(-t )

(Note that we don't find out anything about c, but that's okay since it would have gotten absorbed into the first characteristic solution exp(-t ) anyway.)

• y'' + 4y' + 3y = -(9t + 6) … … … [2]

Compute the derivatives of y₂ :

y₂ = at + b

y₂' = a

y₂'' = 0

Substitute these into [2] :

4a + 3 (at + b) = -9t - 6

3at + 4a + 3b = -9t - 6

→ 3a = -9 and 4a + 3b = -6

→ a = -3 and b = 2

→ y₂ = -3t + 2

Then the general solution to the original ODE is

y(t) = C₁ exp(-t ) + C₂ exp(-3t ) + (2t ² + t ) exp(-t ) - 3t + 2

Use the initial conditions y (0) = 2 and y' (0) = 2 to solve for C₁ and C₂ :

y (0) = C₁ + C₂ + 2 = 2

→ C₁ + C₂ = 0 … … … [3]

y'(t) = -C₁ exp(-t ) - 3C₂ exp(-3t ) + (-2t ² + 3t + 1) exp(-t ) - 3

y' (0) = -C₁ - 3C₂ + 1 - 3 = 2

→ C₁ + 3C₂ = -4 … … … [4]

Solve equations [3] and [4] to get C₁ = 2 and C₂ = -2. Then the particular solution to the initial value problem is

y(t) = -2 exp(-3t ) + (2t ² + t + 2) exp(-t ) - 3t + 2

7 0

3 years ago