Why do all verified people who answer, are always wrong

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Answer 1

Answer:

Answer:

they probably do it just to get more answer points or whatever the brianily thing is called

Step-by-step explanation:

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Solve the system of equations.6x – 3y = 3
–2x + 6y = 14
What number would you multiply the second equation by in order to eliminate the x-terms when adding to the first equation?

What number would you multiply the first equation by in order to eliminate the y-terms when adding to the second equation?

Answers

Keywords:

Systems of equations, variables, values, steps

For this case we have a system of two equations with two variables given by "x" and "y" respectively. We must solve the system by finding the values of the variables. For this, we follow the steps below:

$6x - 3y = 3\ (1)\n-2x + 6y = 14\ (2)$

Step 1:

We multiply the second equation by 3:

$3 * (- 2x + 6y = 14)\n-6x + 18y = 42$

Step 2:

We add both equations:

$6x - 3y = 3\n-6x + 18y = 42\n-6x + 6x-3y + 18y = 42 + 3\n15y = 45$

$y = \frac {45} {15}\ny = 3$

Step 3:

We substitute $y=3$ in the first equation:

$6x - 3 (3) = 3\n6x-9 = 3\n6x = 3 + 9\n6x = 12\nx = \frac {12} {6}x = 2$

Thus, the solution of the system is given by $(x, y) = (2,3)$

Answer:

The second equation must be multiplied by "3" to eliminate the terms of the "x" when added with the first equation

The first equation must be multiplied by "2" to eliminate the terms of the "y" when added with the second equation

The system solution is given by $(x, y) = (2,3)$

Inverse laplace of [(1/s^2)-(48/s^5)]

Answers

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I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform $F(s)$ and $G(s)$ , then $\mathcal{L}^(-1)\{aF(s)+bG(s)\} = a\mathcal{L}^(-1)\{F(s)\}+b\mathcal{L}^(-1)\{G(s)\}$

Given that $(1)/(s^2) = (1!)/(s^2)$ and $-(48)/(s^5) = -2\cdot(4!)/(s^5)$

So you have $\mathcal{L}^(-1)\left\{(1)/(s^2) - 2\cdot(4!)/(s^5)\right\} = \mathcal{L}^(-1)\left\{(1)/(s^2)\right\} - 2\mathcal{L}^(-1)\left\{(4!)/(s^5)\right\}$

From Table of Laplace Transform, you have $\mathcal{L}\{t^n\} = (n!)/(s^(n+1))$ and hence $\mathcal{L}^(-1)\left\{(n!)/(s^(n+1))\right\} = t^n$

So you have $\mathcal{L}^(-1)\left\{(1)/(s^2)\right\} - 2\mathcal{L}^(-1)\left\{(4!)/(s^5)\right\} = \boxed{t-2t^4}$ .

Hope this helps...

Final answer:

To find the inverse Laplace transform of the given expression, use partial fraction decomposition to simplify it into individual fractions and then find their inverse transforms.

Explanation:

To find the inverse Laplace transform of the given expression, we can use partial fraction decomposition. First, we factor the denominator: s2*(s3-48). The next step is to represent the expression as a sum of simpler fractions:

1/s2 - 48/s5 = A/s + B/s2 + C/(s - 2) + D/(s + 2) + E/(s + 4) + F/(s2 - 4)

Next, we solve for A, B, C, D, E, and F by performing algebraic manipulations and equating the corresponding coefficients. Finally, we can look up the inverse Laplace transform of each individual fraction term in tables or by using known formulas.

Learn more about Inverse Laplace Transform here:

brainly.com/question/31322563

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Mai bought a television for $325. The tax rate is 8%. Which proportion can be used to find the amount of tax Mai will pay?

Answers

tax rate = 8%

Thus, tax amount = 8% of 325

= 8/100 x 325

= 26.

Thus, Mai will pay 26$ as tax charge

Please help! Hi, I'm stuck on a part of this question, we have to solve using Gauss Jordan Elimination: A corporation wants to lease a fleet of 12 airplanes with a combined carrying capacity of 220 passengers. The three available types of planes carry 10,15,20 passengers. How many of each type of plane should be leased?

Answers

Do common sense has anything to do with this question? Because
corporation might forget the arbitrary 12 airplane rule
and have 11 times 20 = 220.
But perhaps they get a discount if all 12 hangars are used!

Playing by the rules of the question:-
Let a, b, c be the aeroplanes that carry
10, 15, and 20 passengers, respectively,

a + b + c = 12
10a +15b +20c = 220

We may rearrange these as
4a + 4b + 4c = 48
2a + 3b + 4c = 44
Now subtract
2a + b = 4

There appear to be 3 options
a = 0, b = 4 which means 220 = 0*10 + 4*15 + 8*20
a = 1, b = 2 which means 220 = 1*10 + 2*15 + 9*20
a = 2, b = 0 which means 220 = 2*10 + 0*15 + 7*20

If you dismiss the ones with zeros it may be that
the expected answer to "How many of each plane" is

One that carries 10, 2 that carries 15, and 9 that carries 20

The sum of a number times 2 and 22 is at least 19.

Answers

Frist step is to write it out:

2a + 22 ≤ 19

then you solve for a to find that missing number:

2a ≤ -3
a ≤ -3/2

make sense?

Planes have edges true or false

Answers

False planes do not have edges.

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