(x^3-5x^2+7x-12) (x-4)

Please answer my question bcz I can’t understand

RUDIKE [14]

Please send more context as of the "green numbers".

For the multiplication,

i 49 x 10 = 490
490 ÷ 10 = 49

ii 2.3 ÷ 10 = 0.23
0.23 x 10 = 2.3

iii 0.034 x 1000 = 34
34 ÷ 1000 = 0.034

iv 876 ÷ 100 = 8.76
8.76 x 100 = 876

Hope this helps :)

3 0

3 years ago

Which two tables represent the same function?

topjm [15]

Answer:

The 1st and the 5th tables represent the same function

Step-by-step explanation:

* Lets explain how to solve the problem

- There are five tables of functions, two of them are equal

- To find the two equal function lets find their equations

- The form of the equation of a line whose endpoints are (x1 , y1) and

(x2 , y2) is $\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

* Lets make the equation of each table

# (x1 , y1) = (4 , 8) and (x2 , y2) = (6 , 7)

∵ x1 = 4 , x2 = 6 and y1 = 8 , y2 = 7

∴ $\frac{y-8}{x-4}=\frac{7-8}{6-4}$

∴ $\frac{y-8}{x-4}=\frac{-1}{2}$

- By using cross multiplication

∴ 2(y - 8) = -1(x - 4) ⇒ simplify

∴ 2y - 16 = -x + 4 ⇒ add x and 16 for two sides

∴ x + 2y = 20 ⇒ (1)

# (x1 , y1) = (4 , 5) and (x2 , y2) = (6 , 4)

∵ x1 = 4 , x2 = 6 and y1 = 5 , y2 = 4

∴ $\frac{y-5}{x-4}=\frac{4-5}{6-4}$

∴ $\frac{y-5}{x-4}=\frac{-1}{2}$

- By using cross multiplication

∴ 2(y - 5) = -1(x - 4) ⇒ simplify

∴ 2y - 10 = -x + 4 ⇒ add x and 10 for two sides

∴ x + 2y = 14 ⇒ (2)

# (x1 , y1) = (2 , 8) and (x2 , y2) = (8 , 5)

∵ x1 = 2 , x2 = 8 and y1 = 8 , y2 = 5

∴ $\frac{y-8}{x-2}=\frac{5-8}{8-2}$

∴ $\frac{y-8}{x-2}=\frac{-3}{6}=====\frac{y-8}{x-2}=\frac{-1}{2}$

- By using cross multiplication

∴ 2(y - 8) = -1(x - 2) ⇒ simplify

∴ 2y - 16 = -x + 2 ⇒ add x and 16 for two sides

∴ x + 2y = 18 ⇒ (3)

# (x1 , y1) = (2 , 10) and (x2 , y2) = (6 , 14)

∵ x1 = 2 , x2 = 6 and y1 = 10 , y2 = 14

∴ $\frac{y-10}{x-2}=\frac{14-10}{6-2}$

∴ $\frac{y-10}{x-2}=\frac{4}{4}======\frac{y-10}{x-2}=1$

- By using cross multiplication

∴ (y - 10) = (x - 2)

∴ y - 10 = x - 2 ⇒ add 2 and subtract y in the two sides

∴ -8 = x - y ⇒ switch the two sides

∴ x - y = -8 ⇒ (4)

# (x1 , y1) = (2 , 9) and (x2 , y2) = (8 , 6)

∵ x1 = 2 , x2 = 8 and y1 = 9 , y2 = 6

∴ $\frac{y-9}{x-2}=\frac{6-9}{8-2}$

∴ $\frac{y-9}{x-2}=\frac{-3}{6}======\frac{y-9}{x-2}=\frac{-1}{2}$

- By using cross multiplication

∴ 2(y - 9) = -1(x - 2) ⇒ simplify

∴ 2y - 18 = -x + 2 ⇒ add x and 18 for two sides

∴ x + 2y = 20 ⇒ (5)

- Equations (1) and (5) are the same

∴ The 1st and the 5th tables represent the same function

3 0

3 years ago

Let φ(x, y) = arctan (y/x) .

Alexandra [31]

Answer:

a) $\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

b) $\large \mathbb{R}^2-\{(0,0)\}$

c) the points of the form (x, -x) for x≠0

Step-by-step explanation:

a)

If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be

$\large F(x,y)=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y})$

On one hand we have,

$\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}$

On the other hand,

$\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}$

So

$\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

b)

The domain of definition of F is

$\large \mathbb{R}^2-\{(0,0)\}$

i.e., all the plane X-Y except the (0,0)

c)

Here we want to find all the points such that

$\large (-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})=(k,k)$

where k is a real number other than 0.

But this means

$\large -\frac{y}{x^2+y^2}=\frac{x}{x^2+y^2}\Rightarrow y=-x$

So, all the points in the line y = -x except (0,0) are parallel to the vector field F, that is, the points (x, -x) with x≠ 0

8 0

3 years ago

Which segment is a diameter of this circle? Ο TX O VX Ο XY

professor190 [17]

Answer:

The answer is VX.

Step-by-step explanation:

This is because the line segment VX, reaches from one end of the circle to the other, touch both sides of the circle. The other options do not make sense since segment XY does not go through the center of the circle and TX is the radius of the circle. The radius is half the diameter of the circle.

Therefore, the answer is VX.

5 0

2 years ago

I\ need\ the\ answers\ for\ 6.07\ 8th\ grade\ on\ peak\ fueled

bonufazy [111]

Answer:

what's the question that you need answered?

Step-by-step explanation:

?

6 0

2 years ago