Synthetic Division -2 | 1 2 -3 1

Mathematics

1 answer:

Tom [10]2 years ago

6 0

Answer:

Step-by-step explanation:

I am assuming that the division problem looks like this:

-2| 1 2 -3 1

Going off that assumption, we will work this problem. The first thing you always do in the execution of synthetic division is to bring down the first number. Then multiply that number by the one "outside", which is -2, then put that number up under the next number in the line:

-2| 1 2 -3 1

-2

Now add the 2 and -2 and bring that down as a 0 and multiply the -2 times the 0:

-2| 1 2 -3 1

-2 0

1 0

Now add -3 and 0 to get -3 and multiply that -3 times the -2 and put the product up under the next numbe in line;

-2| 1 2 -3 1

-2 0 6

1 0 -3

Now add the 1 and the 6 to get the remainder:

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Allushta [10]

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3 years ago

E temperature t of a metal sphere is inversely proportional to the distance from the centre of the sphere (the origin (0, 0, 0))

olasank [31]

(-x, -y, -z) the vector form (x, y, z) to the origin, in the direction of the greatest increase.

<h3>How to calculate temperature?</h3>

The quantity or amount of radiation contained in material or item, as measured by a temperature or sensed by touch and stated on a numerical scale.

The temperature T of a ball bearing is inverse to the length first from the origin, which we consider to be the center of the ball. The temperature at the exact location

(x, y, z) is a point on the sphere, the temperature at this point is given by

$T(x,y,z)= \dfrac{k}{(x^2 +y^2+z^2)^{1/2}}$

Where k is constant, then we have

T(1, 2, 2) = k/3 = 170

k = 510

$T(x,y,z)= \dfrac{510}{(x^2 +y^2+z^2)^{1/2}}$

Then we have

$\rm \triangledown T = \left ( -\dfrac{510x}{(x^2 +y^2+z^2)^{1/2}}, -\dfrac{510y}{(x^2 +y^2+z^2)^{1/2}}, \dfrac{510z}{(x^2 +y^2+z^2)^{1/2}} \right )\\\\\\\triangledown T (1, 2, 2) = -\dfrac{510}{27} (1, 2, 2) =-\dfrac{170}{9} (1, 2, 2)$

Then the direction will be from (1, 2, 2) to (4, 3, 5) will be (3, 1, 3)

So u = (3/√19, 1/√19, 3/√19)

Then we get

$\rm \triangledown T \cdot u = \dfrac{-170}{9}(1, 2, 2) \cdot \left (\dfrac{3}{\sqrt{19}}, \dfrac{1}{\sqrt{19}}, \dfrac{3}{\sqrt{19}} \right )\\\\\\\triangledown T \cdot u = - \dfrac{170}{9\sqrt{19}} (3 + 2 + 6)\\\\\\\triangledown T \cdot u = -\dfrac{1870}{9\sqrt{19}}$

The direction of the greatest inverse in the temperature is given by any vector parallel to and having the same direction $\triangledown$ T. (-x, -y, -z) the vector form (x, y, z) to the origin, in the direction of the greatest increase.