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3 years ago

Me pueden ayudar con este problema:(

Mathematics

1 answer:

Korvikt [17]3 years ago

7 0

te sugiero que veas este video para que lo veas y cuando lo veas te explicara como hacerlo

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What is the slope of the line that passes through the points (-4, 8) and (-14, 8) ?

jasenka [17]

Answer:

Step-by-step explanation:

(8-8)/(-14+4) = 0/-10 = 0 is the slope

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3 years ago

Given the a center (-1, -2) and a radius r = 2. Identify the circle.

Tatiana [17]

Answer:

1st option

1st graph has the centre on (-1,-2) and the distance of the circumference from the centre is 2

Answered by GAUTHMATH

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3 years ago

How to simplify 40/100

777dan777 [17]

Answer:

40/100=4/10=2/5 2/5 is the simplest form

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3 years ago

Solve the LP problem. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is

Volgvan

Answer:

(x,y)=(0,7.333)

Step-by-step explanation:

We are required to:

Maximize p = x + 2y subject to

x + 3y ≤ 22
2x + y ≤ 14
x ≥ 0, y ≥ 0.

The graph of the lines are plotted and attached below.

From the graph, the vertices of the feasible region are:

(0,7.333)
(4,6)
(0,0)
(7,0)

At (0,7.333), p=0+2(7.333)=14.666

At (4,6), p=4+2(6)=4+12=16

At (0,0), p=0

At (7,0), p=7+2(0)=7

Since 14.666 is the highest, the maximum point of the feasible region is (0,7.333).

At x=0 and y=7.333, the function p is maximized.

7 0

3 years ago

Find a cubic function with the given zeros.

Fed [463]

Answer:

The correct option is D) $f(x) = x^3 + 2x^2 - 2x - 4$ .

Step-by-step explanation:

Consider the provided cubic function.

We need to find the equation having zeros: Square root of two, negative Square root of two, and -2.

A "zero" of a given function is an input value that produces an output of 0.

Substitute the value of zeros in the provided options to check.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x + 4$ .

$f(x) = x^3 + 2x^2 - 2x + 4\\f(x) = (-2)^3 + 2(-2)^2 - 2(-2) + 4\\f(x) =-8 + 2(4)+4 + 4\\f(x) =8$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 + 2x - 4$ .

$f(x) = x^3 + 2x^2 + 2x - 4\\f(x) = (-2)^3 + 2(-2)^2 + 2(-2) - 4\\f(x) =-8+2(4)-4-4\\f(x) =-8$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 - 2x^2 - 2x - 4$ .

$f(x) = x^3 - 2x^2 - 2x - 4\\f(x) = (-2)^3 - 2(-2)^2 - 2(-2) - 4\\f(x) =-8-8+4-4\\f(x) =-16$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-2)^3+2(-2)^2 - 2(-2) - 4\\f(x) =-8+8+4-4\\f(x) =0$

Now check for other roots as well.

Substitute x=√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (\sqrt{2})^3+2(\sqrt{2})^2 - 2(\sqrt{2}) - 4\\f(x) =2\sqrt{2}+4-2\sqrt{2}-4\\f(x) =0$

Substitute x=-√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-\sqrt{2})^3+2(-\sqrt{2})^2 - 2(-\sqrt{2}) - 4\\f(x) =-2\sqrt{2}+4+2\sqrt{2}-4\\f(x) =0$

Therefore, the option is correct.

8 0

3 years ago