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3 years ago

What is the solution to this system of equations?

Mathematics

1 answer:

mixas84 [53]3 years ago

5 0

Answer:

(-2,4)

Step-by-step explanation:

The solution is the point at which the two lines intersect, (-2,4). That point is the only one that satisfies (works) in both equations:

y = x + 6

4 = -2 + 6

and

y = -0.5x + 3

4 = -0.5(-2) + 3

4 = 1 + 3

====

You can also solve it algebraically:

y = x + 6

y = -0.5x + 3

-0.5x + 3 = x + 6 [Use the value of y from the second equation in the first equation]

-1.5x = 3

x = -2

Use this is y = -2 + 6:

y = 4

(-2,4)

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4 years ago

Jim has $25,000 to invest. Jim buys Bond A which pays 10% interest and Bond B which pays 8% interest. At the end of the year Jim

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3 years ago

List the first twenty counting numbers in the indicated base below. twelve (Only digits 0, 1, 2, 9, A, B are used in base twelve

kicyunya [14]

Answer:

0, 1 , 2, 3, 4, 5, 6, 7, 8, 9, A, B, 10, 11, 12, 13, 14, 15, 16, 17, 18

Step-By-Step Explanation:

i will go from 0 to 20:

0: 0

1: 1

2: 2

3: 3

4: 4

5: 5

6: 6

7: 7

8: 8

9: 9

10: A (10 is another 'digit')

11: B

12: 10 (12 = 1*12^1 + 0* 12^0)

13: 11 (13 = 1*12^1 + 1*12^0)

14: 12 (14 = 1*12^1 + 2*12^0)

15: 13

16: 14

17: 15

18: 16

19: 17

20: 18

Just remember to use the notation A and B after the digit nine, for example

22: 1A

23: 1B

24: 20

In other words, the answer is 0, 1 , 2, 3, 4, 5, 6, 7, 8, 9, A, B, 10, 11, 12, 13, 14, 15, 16, 17, 18

6 0

4 years ago

In general, the eigenvalues of an upper triangular matrix are given by the entries on the diagonal. The same is true for a lower

SCORPION-xisa [38]

Answer:

Verified!

Step-by-step explanation:

Upper or lower triangular matrix does not make any difference in finding eigenvalues because equalizing determinant to zero will lead to the same result.

Let's apply it for 2x2 matix:

$A = \left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0\\0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0\\-b&\lambda-c\end{array}\right]=(\lambda-a)(\lambda-c)=0$

So, eigenvalues are entries on the diagonal because zeros in upper side or lower side vanishes the remaining part and only we have $(\lambda-a)(\lambda-c)$ .

So, eigenvalues are $\lambda_1=a,\:\lambda_2=c$

Let's apply it for 3x3 matrix:

$A = \left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0&0\\-b&\lambda-c&0\\-d&-e&\lambda-f\end{array}\right]=(\lambda-a)(\lambda-c)(\lambda-f)=0$

So as above, eigenvalues are entries on the diagonal because zeros in upper side or lower side vanishes the remaining part and only we have $(\lambda-a)(\lambda-c)(\lambda-f)$ .

So eigenvalues are $\lambda_1=a,\:\lambda_2=c,\:\lambda_3=f$

4 0

4 years ago

The mean of a,b,c,and d is 9 and the mean of a,b,c,d,e and f is 12. find the mean of e and f

Simora [160]

Answer:

mean of e + f = 18

Step-by-step explanation:

(a + b + c + d)/4 = 9 Multiply by 4

a + b + c + d = 9 * 4 = 36

(a + b + c + d + e + f ) / 6 = 12

but a + b + c + d = 36

(36 + e + f)/6 = 12 Multiply by 6

36 + e + f = 72 Subtract 36 from both sides

e + f = 36 Divide by 2 to find the mean.

(e + f)/2 = 36/2

mean of e + f = 18

8 0

4 years ago