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3 years ago

The sum of the roots of 2x 2 + 8x - 3 = 0 is -3/2 3/2 -4 4

Mathematics

1 answer:

Sophie [7]3 years ago

7 0

Answer:

-4

Step-by-step explanation:

From the qquadratic formula

x1 = [ -b +√(b² - 4ac) ] / 2a

x2 = [ -b -√(b² - 4ac) ] / 2a

When added together the ± square root cancels out leaving

(-b/2a) + (-b/2a) = -b/a

For 2x² + 8x -3 = 0

-b/a = -8/2

-b/a = -4

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CAN SOMEONE PLEASE HELP. SEGMENTS GEOMETRY

olga55 [171]

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4 years ago

The coordinate plane below represents a city. Points A through F are schools in the city. graph of coordinate plane. Point A is

Firlakuza [10]

Part A;
There are many system of inequalities that can be created such that only contain points C and F in the overlapping shaded regions.

Any system of inequalities which is satisfied by (2, 2) and (3, 4) but is not stisfied by <span>(-3, -4), (-4, 3), (1, -2) and (5, -4) can serve.

An example of such system of equation is
x > 0
y > 0
The system of equation above represent all the points in the first quadrant of the coordinate system.
The area above the x-axis and to the right of the y-axis is shaded.

Part 2:
It can be verified that points C and F are solutions to the system of inequalities above by substituting the coordinates of points C and F into the system of equations and see whether they are true.

Substituting C(2, 2) into the system we have:
2 > 0
2 > 0
as can be seen the two inequalities above are true, hence point C is a solution to the set of inequalities.

Part C:
Given that </span><span>Natalie can only attend a school in her designated zone and that Natalie's zone is defined by y < −2x + 2.

To identify the schools that Natalie is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining Natalie's zone.

For point A(-3, -4): -4 < -2(-3) + 2; -4 < 6 + 2; -4 < 8 which is true

For point B(-4, 3): 3 < -2(-4) + 2; 3 < 8 + 2; 3 < 10 which is true

For point C(2, 2): 2 < -2(2) + 2; 2 < -4 + 2; 2 < -2 which is false

For point D(1, -2): -2 < -2(1) + 2; -2 < -2 + 2; -2 < 0 which is true

For point E(5, -4): -4 < -2(5) + 2; -4 < -10 + 2; -4 < -8 which is false

For point F(3, 4): 4 < -2(3) + 2; 4 < -6 + 2; 4 < -4 which is false

Therefore, the schools that Natalie is allowed to attend are the schools at point A, B and D.
</span>

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4 years ago

The LCD for 3/4 and 1/6 is

OleMash [197]

1/2 im pretty sure im right

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4 years ago

Sarah has 4 more $50 notes than $10 notes in her wallet.If the total amount of the notes is $380,how many $10 notes does Sarah h

Anni [7]

Answer:

Step-by-step explanation:

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2 years ago

Calculate all four second-order partial derivatives and confirm that the mixed partials are equal. f(x,y)= 2e^xy

nadya68 [22]

ANSWER TO QUESTION 1

The given function is

$f(x,y)=2 {e}^{xy}$

The partial derivative of f with respect to x means we are treating y as a constant. The first derivative is

$f_{x} = 2y {e}^{xy}$

and the second derivative with respect to x is,

$f_{xx} = 2 {y}^{2} {e}^{xy}$

ANSWER TO QUESTION 2

The given function is

$f(x,y)=2 {e}^{xy}$

The partial derivative of f with respect to y means we are treating x as a constant. The first derivative is

$f_{y} = 2x{e}^{xy}$

and the second derivative with respect to y is

$f_{yy} = 2 {x}^{2} {e}^{xy}$

ANSWER TO QUESTION 3

Our first mixed partial is

$f_{xy}$

We need to differentiate
$f_{x} = 2y {e}^{xy}$
again. But this time with respect to y.

Since this is a product of two functions of y, we apply the product rule of differentiation to obtain,

$f_{xy} = 2y( {e}^{xy})' + ({e}^{xy})(2y)'$

$f_{xy} = 2xy {e}^{xy} + 2{e}^{xy}$

ANSWER TO QUESTION 4

The second mixed partial is

$f_{yx}$

We need to differentiate
$f_{y} = 2x{e}^{xy}$

again. But this time with respect to x.

Since this is a product of two functions of x, we apply the product rule of differentiation to obtain,

$f_{yx} = 2x({e}^{xy})' + ({e}^{xy})(2x)'$

$f_{yx} = 2xy {e}^{xy} + 2{e}^{xy}$

Hence,

$f_{xy} = 2xy {e}^{xy} + 2{e}^{xy} =f_{yx}$

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3 years ago