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4 years ago

The value of the expression - 3. () is (Do without a calculator and use "T" as your fraction bar.)

Mathematics

1 answer:

gayaneshka [121]4 years ago

3 0

Answer:

-10/21

Step-by-step explanation:

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Plz help please<br><br> Is the relation shown in the arrow diagram a function?

SSSSS [86.1K]

Answer:

The answer is A

Step-by-step explanation:

In order for it to be a function every x value has to have a different y value or every input has to have a different output.

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3 years ago

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Viefleur [7K]

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4 years ago

Y = x and y =x^2 - 2

maxonik [38]

Answer:

(-1, -1) and (2, 2)

Step-by-step explanation:

Substitute for y in the first equation and solve the quadratic.

x = x^2 -2

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(x -2)(x +1) = 0 . . . .factor

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(x, y) = (-1, -1) and (2, 2)

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4 years ago

ANWNSWER THIS FAST PLZZZZ!!!!!!!!!!!!!

devlian [24]

Step-by-step explanation:

$\frac{1}{15}$

0.0666666666666

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3 years ago

Describe the steps required to determine the equation of a quadratic function given its zeros and a point.

faltersainse [42]

Answer:

Procedure:

1) Form a system of 3 linear equations based on the two zeroes and a point.

2) Solve the resulting system by analytical methods.

3) Substitute all coefficients.

Step-by-step explanation:

A quadratic function is a polynomial of the form:

$y = a\cdot x^{2}+b\cdot x + c$ (1)

Where:

$x$ - Independent variable.

$y$ - Dependent variable.

$a$ , $b$ , $c$ - Coefficients.

A value of $x$ is a zero of the quadratic function if and only if $y = 0$ . By Fundamental Theorem of Algebra, quadratic functions with real coefficients may have two real solutions. We know the following three points: $A(x,y) = (r_{1}, 0)$ , $B(x,y) = (r_{2},0)$ and $C(x,y) = (x,y)$

Based on such information, we form the following system of linear equations:

$a\cdot r_{1}^{2}+b\cdot r_{1} + c = 0$ (2)

$a\cdot r_{2}^{2}+b\cdot r_{2} + c = 0$ (3)

$a\cdot x^{2} + b\cdot x + c = y$ (4)

There are several forms of solving the system of equations. We decide to solve for all coefficients by determinants:

$a = \frac{\left|\begin{array}{ccc}0&r_{1}&1\\0&r_{2}&1\\y&x&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$a = \frac{y\cdot r_{1}-y\cdot r_{2}}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x+x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$a = \frac{y\cdot (r_{1}-r_{2})}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$b = \frac{\left|\begin{array}{ccc}r_{1}^{2}&0&1\\r_{2}^{2}&0&1\\x^{2}&y&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$b = \frac{(r_{2}^{2}-r_{1}^{2})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$c = \frac{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&0\\r_{2}^{2}&r_{2}&0\\x^{2}&x&y\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$c = \frac{(r_{1}^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x + x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

And finally we obtain the equation of the quadratic function given two zeroes and a point.

6 0

3 years ago

The value of the expression - 3. () is (Do without a calculator and use "T" as your fraction bar.)​

The value of the expression - 3. () is (Do without a calculator and use "T" as your fraction bar.)