Ask question

Ask question

All categories

3 years ago

6

2 x 9 = ? and also dont you just loooooove my profile picture

2 answers:

hjlf3 years ago

6 0

Answer:

18

Step-by-step explanation:

Verizon [17]3 years ago

5 0

Answer:

18

Step-by-step explanation:

18

2 * 9 = 18

9 + 9 = 18

You might be interested in

6x1/4= ? (fractions)<br><br> 3x2/3 = ?

son4ous [18]

6x1/4=6/4=2/3
3x2/3=6/3=2

4 0

2 years ago

Read 2 more answers

One solution of x^2-64=0 is 8. what is the other solution?

ivann1987 [24]

X² - 64 = 0
(x - 8)(x + 8) = 0

x = - 8, 8

Your other solution is - 8

4 0

3 years ago

Read 2 more answers

When two lines intersect to form two angles, what conjecture can you make about the pairs of angles that are next to each other?

gayaneshka [121]

The pair of angles are adjacent angles and supplementary and their measure =180

7 0

3 years ago

What is the reciprocal of 6/5? <br> A. 1 <br> B. 1 1/5 <br> C. 12/10 <br> D. 5/6

Sergio039 [100]

The reciprocal of 6/5 is D. 5/6

Reciprocal simply means swapping the position of the numbers in the fraction. The numerator becomes the denominator and the denominator becomes the numerator.

We need to get reciprocal of a fraction when division is performed.

For example: 2 ÷ 1/5

2 may be a whole number but in fraction form it is 2/1.

1st fraction = 2/1
2nd fraction = 1/5

In dividing fractions, the 1st step we need to do is to get the reciprocal of the 2nd fraction.

1/5 ⇒ 5/1

Then, we multiply the 1st fraction to the reciprocal of the 2nd fraction.
2/1 * 5/1 = 10

So, 2 ÷ 1/5 = 10

8 0

3 years ago

Read 2 more answers

Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

WINSTONCH [101]

Answer:

a) $y(t) = y_{0}e^{4t} + 2$ . It does not have a steady state

b) $y(t) = y_{0}e^{-4t} + 2$ . It has a steady state.

Step-by-step explanation:

a) $y' -4y = -8$

The first step is finding $y_{n}(t)$ . So:

$y' - 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r - 4 = 0$

$r = 4$

So:

$y_{n}(t) = y_{0}e^{4t}$

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' -4(y_{p}) = -8$

$(C)' - 4C = -8$

C is a constant, so (C)' = 0.

$-4C = -8$

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{4t} + 2$

b) $y' +4y = 8$

The first step is finding $y_{n}(t)$ . So:

$y' + 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r + 4 =$

$r = -4$

So:

$y_{n}(t) = y_{0}e^{-4t}$

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' +4(y_{p}) = 8$

$(C)' + 4C = 8$

C is a constant, so (C)' = 0.

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{-4t} + 2$

6 0

3 years ago