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KiRa [710]
2 years ago
13

In the plane figure, only downward motion (movement leaving you relatively lower than where you were) is allowed. Find the total

number of paths from A to B.

Mathematics
1 answer:
bixtya [17]2 years ago
8 0

<em>Plane figures</em> are shapes with <u>straight </u>sides. Considering the <em>movement</em> to be <em>lower</em> than the initial path, there are eleven (11) <u>paths</u> from A to B.

A <em>plane figure</em> is given shape with <u>straight</u> sides. And these <em>sides </em><u>connect</u> each edge of the figure one to another. Examples are triangles, quadrilaterals, polygons, etc.

A path is a <em>connection</em> or <u>link</u> between two <em>reference </em>points. This could be a <u>footpath</u> or a <u>motorable</u> path or <u>other</u> forms of the path.

Considering the given <em>plane figure</em> in the question, it can be observed that the <u>total</u> <u>number</u> of paths from A to B with respect to the given condition is eleven.

For more clarifications on paths around a given plane figure, visit: brainly.com/question/4871149

#SPJ1

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Answer:

-5/ (x^19 * y^12)

Step-by-step explanation:

-15x^-10 * y ^ -3 * x^-9 * y^-9/3 (It's writing it like this.)

15 and 3 have a GCF of 3.

-5x^-10 * y^-3 * x^-9 * y^-9/1

Reduce similar occurrences. *twice in a row*

-5x^(-10 + (-9)) * y^(-3 + (-9))/1

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Remove extraneous 1.

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Add integers.

-5x^-19 * y^-12

Negative powers change multiplication to division.

-5 / (x^19 * y^12) is the final answer.

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The polynomial function y = x3 -3x2 + 16x - 48 has only one non-repeated x-intercept. What do you know about the complex zeros o
earnstyle [38]
• First way to solve:
We'll manipulate the expression of the equation:

y=x^3-3x^2+16x-48\\\\ y=x^2(x-3)+16(x-3)\\\\ y=(x-3)(x^2+16)

If we have y=0:

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Then, the function has one real zero (x=3) and two imaginary zeros (4i and -4i).

Answer: B


• Second way to solve:

The degree of the function is 3. So, the function has 3 complex zeros.

Since the coefficients of the function are reals, the imaginary roots are in a even number (a imaginary number and its conjugated)

The function "has only one non-repeated x-intercept", then there is only one real zero.

The number of zeros is 3 and there is 1 real zero. So, there are 2 imaginary zeros.

Answer: B.
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