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2 years ago

What is |Ay| in the system of linear equations below? -8x+y=6 3x-2y=-1

2 answers:

6 0

$A=\left[\begin{array}{ccc}-8&1\\3&-2\end{array}\right]\\\\|A_y|=\left|\begin{array}{ccc}-8&6\\3&-1\end{array}\right|=(-8)(-1)-(3)(6)=8-18=-10$

sergiy2304 [10]2 years ago

4 0

Answer:

|Ay| = - 10

y = 10/19

x = 2 - 10/57

Step-by-step explanation:

we have to set the data into array format to be able to solve the problem.

$A=\left[\begin{array}{ccc}-8&1\\3&-2\end{array}\right]\\\\|A_y|=\left|\begin{array}{ccc}-8&6\\3&-1\end{array}\right|=(-8)(-1)-(3)(6)=8-18=-10$

but then to calculate for the values of x and y we solve simultaneously

-24x + 3y = 18

-24x - 16y = 8

19y =10

y = 10/19

substituting the value of y

3x + 10/19 = 6

3x = 6-10/19

x = 2 - 10/57

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Answer:

$f''(x)=36x^2+4$

Step-by-step explanation:

Let's start by finding the first derivative of $f(x)= 3x^4+2x^2-8x+4$ . We can do so by using the power rule for derivatives.

The power rule states that:

$\frac{d}{dx} (x^n) = n \times x^n^-^1$

This means that if you are taking the derivative of a function with powers, you can bring the power down and multiply it with the coefficient, then reduce the power by 1.

Another rule that we need to note is that the derivative of a constant is 0.

Let's apply the power rule to the function f(x).

$\frac{d}{dx} (3x^4+2x^2-8x+4)$

Bring the exponent down and multiply it with the coefficient. Then, reduce the power by 1.

$\frac{d}{dx} (3x^4+2x^2-8x+4) = ((4)3x^4^-^1+(2)2x^2^-^1-(1)8x^1^-^1+(0)4)$

Simplify the equation.

$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x^1-8x^0+0)$
$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x-8(1)+0)$

$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x-8)$
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Now, this is only the first derivative of the function f(x). Let's find the second derivative by applying the power rule once again, but this time to the first derivative, f'(x).