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3 years ago

Factorize: 2ax-6ay+bx-3by

Mathematics

2 answers:

kotykmax [81]3 years ago

4 0

Answer:

Step-by-step explanation:

2a(x - 3y) + b(x - 3y)

Elodia [21]3 years ago

3 0

Answer:

2ax − 6ay + bx − 3

Step-by-step explanation:

The expression is not factorable with rational numbers

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2 years ago

Why is it helpful to rewrite inequalities

mr_godi [17]

Answer:

that puts the solution in the form ...

variable is ...

Step-by-step explanation:

It isn't always.

Often, we like to have a solution be in the form ...

variable is ...

So, for an inequality, that puts the variable on the left:

x > 3

y < 27

Personally, I like to see the answer in a form that has the variable and its values in the same relation as on a number line. This means, my preferred inequality symbols are < or ≤, since those have the smaller numbers on the left. I would write the first example above as ...

3 < x

showing that the shaded portion of the number line (representing possible values of the variable) is to the right of the open circle at 3. For me, it is more mental effort to translate x > 3 to the same image.

The forms we choose to use are all about making communication as easy as possible.

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3 years ago

Taylor writes an expression with 5 terms. All 5 terms are like terms. How many terms are in the equivalent expression with the l

Volgvan

I have no clue I’m dum

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2 years ago

The factors for x2 + 5x + 6 are x + 2 and x + 3.

Nookie1986 [14]

It is true that (x+2)(x+3) = x2 +5x + 6

5 0

3 years ago

Which of the following equations represents the perpendicular bisector of WX graphed below?

kotykmax [81]

The coordinates of the 2 given points are W(-5, 2), and X(5, -4).

First, we find the midpoint M using the midpoint formula:

$\displaystyle{ M_{WX}= (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} )= (\frac{-5+5}{2}, \frac{2+(-4)}{2} )=(0, -1).$

Nex, we find the slope of the line containing M, perpendicular to WX. We know that if m and n are the slopes of 2 parallel lines, then mn=-1.

The slope of WX is $\displaystyle{ m= \frac{y_2-y_1}{x_2-x_1}= \frac{2-(-4)}{-5-5}= \frac{6}{-10}= -\frac{3}{5}$ .

Thus, the slope n of the perpendicular line is $\displaystyle{ \frac{5}{3}$ .

The equation of the line with slope $\displaystyle{ n= \frac{5}{3}$ containing the point M(0, -1) is given by:

$\displaystyle{ y-(-1)=\frac{5}{3}(x-0)$

$\displaystyle{ y+1= \frac{5}{3}x$

$\displaystyle{ 3y+3=5x$

$\displaystyle{ 5x-3y-3=0$

Answer: 5x-3y-3=0

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3 years ago