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Ber [7]
3 years ago
6

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

Mathematics
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

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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
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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
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