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2 years ago

X + 2y = 5 x - 3y = 7 What is the value of the y-determinant? -2 -1 2

Mathematics

1 answer:

grandymaker [24]2 years ago

5 0

Answer:

y-determinant = 2

Step-by-step explanation:

Given the following system of equation:

x + 2y = 5
x - 3y = 7

Let's represent it using a matrix:

$\left[\begin{array}{ccc}1&2\\1&-3\end{array}\right] = \left[\begin{array}{ccc}5\\7\end{array}\right]$

The y‐numerator determinant is formed by taking the constant terms from the system and placing them in the y‐coefficient positions and retaining the x‐coefficients. Then:

$\left[\begin{array}{ccc}1&5\\1&7\end{array}\right]$

y-determinant = (1)(7) - (5)(1) = 2.

Therefore, the y-determinant = 2

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2 years ago

What is the value of \dfrac{d}{dx}\left(\dfrac{2x+3}{3x^2-4}\right) dx d ( 3x 2 −4 2x+3 )start fraction, d, divided by, d, x

stellarik [79]

Answer:

Step-by-step explanation:

We are asked to find the value of expression $\frac{d}{dx}(\frac{2x+3}{3x^2-4})$ at $x=-1$ .

First of all, we will find the derivative of the given expression using "Quotient Rule of Derivatives" as shown below:

$(\frac{f(x)}{g(x)})'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{(g(x))^2}$

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

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

$\frac{2*(3x^2-4)-(2x+3)*(6x)}{(3x^2-4)^2}$

$\frac{6x^2-8-12x^2-18x}{(3x^2-4)^2}$

$\frac{-6x^2-18x-8}{(3x^2-4)^2}$

Therefore, our required derivative is $\frac{-6x^2-18x-8}{(3x^2-4)^2}$ .

Now, we will substitute $x=-1$ in our derivative to find the required value as:

$\frac{-6(-1)^2-18(-1)-8}{(3(-1)^2-4)^2}$

$\frac{-6(1)+18-8}{(3(1)-4)^2}$

$\frac{-6+18-8}{(3-4)^2}$

$\frac{4}{(-1)^2}$

$\frac{4}{1}$

$4$

Therefore, the value of expression $\frac{d}{dx}(\frac{2x+3}{3x^2-4})$ at $x=-1$ is 4.

6 0

2 years ago

A number is selected, at random, from the set {1,2,3,4,5,6,7,8}.

Olegator [25]

Applying the formula, you have:

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B = the number is odd

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$P(A) = \dfrac{4}{8} = \dfrac{1}{2}$

because 4 out of 8 numbers are prime: 2, 3, 5 and 7.

Similarly, we have

$P(B) = \dfrac{4}{8} = \dfrac{1}{2}$

because 4 out of 8 numbers are odd: 1, 3, 5 and 7.

Finally,

$P(A \land B) = \dfrac{3}{8}$

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So, applying the formula, we have

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Note:

I think that it is important to have a clear understanding of what's happening from a conceptual point of you: conditional probability simply changes the space you're working with: you are not asking "what is the probability that a random number, taken from 1 to 8, is prime?"

Rather, you are adding a bit of information, because you are asking "what is the probability that a random number, taken from 1 to 8, is prime, knowing that it's odd?"

So, we're not working anymore with the space {1,2,3,4,5,6,7,8}, but rather with {1,3,5,7} (we already know that our number is odd).

Out of these 4 odd numbers, 3 are primes. This is why the probability of picking a prime number among the odd numbers in {1,2,3,4,5,6,7,8} is 3/4: they are literally 3 out of 4.

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3 years ago