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3 years ago

12

(6m³n)²(-4m⁵n) must be simplified

1 answer:

Alekssandra [29.7K]3 years ago

8 0

Answer:

-144m^11n^3

Step-by-step explanation:

-(6m^3)^2 x 4m^5n

-36m^6n^2 x 4m^5n

Answer:

-144m^11n^3

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Help me out with this

lutik1710 [3]

The process is similar to (and easier* than) adding three 4-digit numbers. Add the numbers in each column.

The sum is ...
(2+1-3)x³ +(-4+6+2)x² +(6-8-4)x +(-3+12-7)
= 0x³ +4x² -6x +2

The sum is 4x² -6x +2

_____
* The process is easier because there are no "carry" operations from one column to another as there may be when adding multi-digit numbers.

3 0

3 years ago

Line Segment BC has endpoints B (3,5) and C (7,15). Find the missing coordinates of A (x,9) and D (17,y) such that AB and CD are

o-na [289]

Answer:

<em>x=-7</em>

<em>y=11</em>

Step-by-step explanation:

<u>Perpendicular Vectors</u>

Two vectors defined as their endpoints $\vec u=$ and $\vec v=$ are perpendicular if their dot product a.b is zero. The dot product is

$\vec u.\vec v=ac+bd$

In other words

ac+bd=0

Let's treat all the points as the extremes of vectors, so we can easily find the missing coordinates

B (3,5) and C (7,15) define a segment, the vector

$\overrightarrow{BC}==$

The point A is A (x,9), we need to form a vector with B

$\overrightarrow{AB}==$

this vector must be perpendicular to BC, so, applying the dot product we have

$4(x-3)+40=0$

$4x-12=-40$

$x=-7$

The point D is D(17,y), we need to form a vector with C

$\overrightarrow{CD}==$

this vector must be perpendicular to BC, so, applying the dot product we have

$(4)(10)+(10)(y-15)=0$

$40+10y-150=0$

$10y=110$

$y=11$

The points are

$\boxed{A(-7,9), D(17,11)}$

4 0

3 years ago

Write an equation in slope-intercept form of the line that passes through (6,-2) and (12,1)

yarga [219]

Equation in slope-intercept form of the line that passes through (6,-2) and (12,1) is:

$y =\frac{1}{2}x-5$

Step-by-step explanation:

Given points are:

(x1,y1) = (6,-2)

(x2,y2) = (12,1)

The slope intercept form is:

$y=mx+b$

We have to find the slope first

$m =\frac{y_2-y_1}{x_2-x_1}\\=\frac{1-(-2)}{12-6}\\= \frac{1+2}{6}\\=\frac{3}{6}\\=\frac{1}{2}$

Putting the value of slope

$y = \frac{1}{2}x+b$

To find the value of b, putting (12,1) in the equation

$1 = \frac{1}{2}(12)+b\\1 = 6+b\\b = 1-6\\b=-5$

Putting the values of m and b

$y =\frac{1}{2}x-5$

Hence,

Equation in slope-intercept form of the line that passes through (6,-2) and (12,1) is:

$y =\frac{1}{2}x-5$

Keywords: Equation of line, slope-intercept form

Learn more about equation of line at:

#LearnwithBrainly

8 0

3 years ago

F10<br> Find the gradient of the line segment between the points (-2,-2) and (0,-6).<br> 2

Pepsi [2]

Answer:

-2

Step-by-step explanation:

M = (y - yo/x - xo)

(-6 + 2) / (0 + 2)

-4 / -2

= - 2

3 0

3 years ago

I WILL MARK BRAINLIST

Vladimir79 [104]

Answer:

a

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers