All categories

2 years ago

!!HELP!! !!PLS!!

Mathematics

2 answers:

creativ13 [48]2 years ago

8 0

First part

$6x + blank = 10\\blank = 10 - 6x$

Thus the first part's blank should be filled in with 4x

Second part

$6x-(blank)= 4x -10\\-(blank) =4x - 10 - 6x\\-(blank) = -2x -10\\blank = 2x + 10$

Thus the second part's blank should be filled in with 2x + 10

Hope that helps!

Gala2k [10]2 years ago

4 0

Fist part 4x

Second part 2x+10

You might be interested in

What is the point-slope form of a line with slope 3 that contains the point (2,1)

Ilya [14]

<h3>Answer: y - 1 = 3(x - 2)</h3>

Explanation:

The point-slope form is this general template

y - y1 = m(x - x1)

We replace m with the given slope 3, so we get to this

y - y1 = 3(x - x1)

Then replace the given point (x1, y1) with (2, 1) to end up with

y - 1 = 3(x - 2)

4 0

2 years ago

Solve the inequalities -150x ≥ − 2,400 and -336 ≥ − 21y.

astra-53 [7]

Ok so easy peasy
remember that when divide or multiply by negative, reverse the inequality symbol example
2>3 times -1=
-2<-3 so

-150x<u>></u>-2400
divide both sides by -150
flip sign
<u />x<u><</u>16

-336<u>></u>-21y
divide both sides by -21
flip sign
16<u><</u>y

they seem to have the same solution

x=y<u>></u>16

4 0

3 years ago

Read 2 more answers

1/2-1/4<br><br> help?:) uhhhh thankssss

sertanlavr [38]

1/2 - 1/4 = 2/4 - 1/4 = 1/4

I hope this helps!

4 0

2 years ago

5. Find the value(s) of x so that the line containing the points (2x + 3, x + 2) and (0, 2) is

Dvinal [7]

Answer:

x = -2 or -9

Step-by-step explanation:

You want the values of x such that the line defined by the two points (2x+3, x+2) and (0, 2) is perpendicular to the line defined by the two points (x+2, -3-3x) and (8, -1).

<h3>Slope</h3>

The slope of a line is given by the slope formula:

m = (y2 -y1)/(x2 -x1)

Using the formula, the slopes of the two lines are ...

m1 = (2 -(x+2))/(0 -(2x+3)) = (-x)/(-2x-3) = x/(2x +3)

and

m2 = (-1 -(-3-3x))/(8 -(x+2)) = (2+3x)/(6 -x)

<h3>Perpendicular lines</h3>

The slopes of perpendicular lines have product of -1:

$\dfrac{x}{2x+3}\cdot\dfrac{2+3x}{6-x}=-1\\\\x(3x+2)=(2x+3)(x-6)\qquad\text{multiply by $(2x+3)(6-x)$}\\\\3x^2+2x=2x^2-9x-18\qquad\text{eliminate parentheses}\\\\x^2+11x+18=0\qquad\text{put in standard form}\\\\(x+2)(x+9)=0\qquad\text{factor}$

<h3>Solutions</h3>

The values of x that satisfy this equation are x = -2 and x = -9. The attached graphs show the lines for each of these cases.

4 0

1 year ago

Wha are the equations. of the asymptotes of the graph of the function f(x)=3x^2-2x-1/x^2+3x+10

Leviafan [203]

The asymptote is the line the graph comes close to but never touches, so if you graph it out you should be able to see it! Sorry I couldn't help any more but good luck

3 0

3 years ago