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3 years ago

Solve the system of linear equations.

Mathematics

1 answer:

icang [17]3 years ago

7 0

Answer: The set does not have a solution

Step-by-step explanation:

Adding Equations 1 & 3 we get 5x = 7. This gives x = 7/5

Putting this value of x in eq. 2 we get

-2y + 2z = -1-(7/5) or

2y - 2z = 12/5 or 5y - 5z = 6

Multiplying eq. 1 by 2 we get

4x + 2y - 2z = 6

adding this with eq. 2 we get 5x = 5 or x = 1

As the common solution for x from equations 1&3 does not satisfy eq. 1&2 it comes out that the three equations do not have a common solution.

Same can be verified by using different sets of two equations also.

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Find the Laplace Transform of the following function. <br><br> f(t) = 10 cos (12t+ 60°) u(t)

allsm [11]

Answer:

L(f(t)) = $\frac{5s - 60\sqrt{3} }{s^2 + 144}$

Step-by-step explanation:

f ( t ) = 10 cos ( 12t + 60°) u(t)

attached below is a detailed solution of the given problem

L(f(t)) = $\frac{5s - 60\sqrt{3} }{s^2 + 144}$

8 0

2 years ago

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erma4kov [3.2K]

Answer:

Yes below

Step-by-step explanation:

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y=12m+64.50

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3 years ago

How do you make a fraction that’s greater than one or just a fraction into a mixed number?

sergiy2304 [10]

Answer:

You multiply the denominator and the whole number then add the numerator.

Step-by-step explanation:

Example: $3\frac{2}{4}$

You multiply 3 and 4

3*4=12

Then add the numerator

12+2=14

The final answer would be

$\frac{14}{4}$

6 0

3 years ago

In a class of 19 students, 3 are math majors. A group of four students is chosen at random. (Round your answers to four decimal

KatRina [158]

Answer:

(a) The probability is 0.4696

(b) The probability is 0.5304

Step-by-step explanation:

The total number of ways in which we can select k elements from a group n elements is calculate as:

$nCx=\frac{n!}{x!(n-x)!}$

So, the number of ways in which we can select four students from a group of 19 students is:

$19C4=\frac{19!}{4!(19-4)!}=3,876$

On the other hand, the number of ways in which we can select four students with no math majors is:

$(16C4)*(3C0)=(\frac{16!}{4!(16-4)!})*(\frac{3!}{0!(3-0)!})=1820$

Because, we are going to select 4 students form the 16 students that aren't math majors and select 0 students from the 3 students that are majors.

At the same way, the number of ways in which we can select four students with one, two and three math majors are 1680, 360 and 16 respectively, and they are calculated as:

$(16C3)*(3C1)=(\frac{16!}{3!(16-3)!})*(\frac{3!}{1!(3-1)!})=1680$

$(16C2)*(3C2)=(\frac{16!}{2!(16-2)!})*(\frac{3!}{2!(3-1)!})=360$

$(16C1)*(3C3)=(\frac{16!}{1!(16-1)!})*(\frac{3!}{3!(3-3)!})=16$

Then, the probability that the group has no math majors is:

$P=\frac{1820}{3876} =0.4696$

The probability that the group has at least one math major is:

$P=\frac{1680+360+16}{3876} =0.5304$

The probability that the group has exactly two math majors is:

$P=\frac{360}{3876} =0.0929$

6 0

3 years ago

Find the area of the figure.

Margarita [4]

Answer:

60 ft^2

Step-by-step explanation:

6 0

2 years ago

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