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3 years ago

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The system of equations below is consistents with infinity many solutions where a and b are constants.Find the values of a and b

. 2x+y=6 ax+by=9

1 answer:

Kitty [74]3 years ago

7 0

Answer:

The values of $a$ and $b$ so that the two linear equations have infinite solutions are $3$ and $\frac{3}{2}$ , respectively.

Step-by-step explanation:

Two linear equations with two variables have infinite solution if and only if they are<em> linearly dependent</em>. That is, one linear equation is a multiple of the other one. Let be the following system of linear equations:

$2\cdot x + y = 6$ (1)

$a\cdot x + b\cdot y = 9$ (2)

The following condition must be observed:

$r = \frac{a}{2} = \frac{b}{1} = \frac{9}{6}$ (3)

After some quick operations, we find the following information:

$r = \frac{3}{2}$ , $a = 3$ , $b = \frac{3}{2}$

The values of $a$ and $b$ so that the two linear equations have infinite solutions are $3$ and $\frac{3}{2}$ , respectively.

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Find the 2nd Derivative:<br> f(x) = 3x⁴ + 2x² - 8x + 4

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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.

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

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

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).

$\frac{d}{d} (f'x) = \frac{d}{dx} (12x^3+4x-8)$

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

Simplify the equation.

$\frac{d}{dx} (12x^3+4x-8) = (36x^2 + 4x^0 - 0)$
$\frac{d}{dx} (12x^3+4x-8) = (36x^2 + 4(1) - 0)$
$\frac{d}{dx} (12x^3+4x-8) = (36x^2 + 4 )$

Therefore, this is the 2nd derivative of the function f(x).

We can say that: $f''(x)=36x^2+4$

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2 years ago

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If P = (-2,7), Find:<br> Ry-axis (P)<br> ([?], [])

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Answer:

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Step-by-step explanation:

$R_{\text{y-axis}}(\text{P})$ represents the reflection of a point P about the y-axis.

If a point (x, y) is reflected across y-axis, rule for the reflection is,

(x, y) → (-x, y)

If a point (-2, 7) follows the same rule, coordinates of the image point will be,

P(-2, 7) → P'(2, 7)

Therefore, answer is (2, 7).

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