C+ax=dxthis one has all variables and no numbers

1 answer:

6 0

That's OK, but you have not said which variable you want to solve it for.

To solve for 'x':
 c + ax = dx

Subtract c from each side: ax = dx - c

Subtract dx from each side: ax - dx = -c

Factor the left side: x (a - d) = -c

Divide each side by (a - d) : x = -c / (a - d) or x = c / (d - a) .

To solve for 'c': 
 c + ax = dx

Subtract ax from each side and factor: c = x (d - a) 

To solve for 'd': 
 c + ax = dx

Divide each side by 'x': d = c/x + a .

To solve for 'a':
 c + ax = dx

Subtract 'c' from each side: ax = dx - c

Divide each side by 'x': a = d - c/x .
.

You might be interested in

HELP PLSS ANYONE? I'VE TO SUBMIT ASAP

Alenkinab [10]

x =14

Answer:

$m\angle ABC = 51°$

Step-by-step explanation:

11. By inscribed angle theorem:

$83°=\frac{1}{2} \times (9x+40)°\\\\83°\times 2 = (9x+40)°\\166°= (9x+40)°\\166 = 9x + 40\\166 - 40 = 9x\\126 = 9x\\\\x = \frac{126}{9}\\\\\huge \red {\boxed {x = 14}}$

12. Again by inscribed angle theorem:

$(5x+11)°=\frac{1}{2} \times (16x-26)°\\\\2\times (5x+11)°= (16x-26)°\\(10x+22)°= (16x-26)°\\10x + 22 = 16x - 26\\22+26 = 16x - 10x\\48= 6x\\\\x = \frac{48}{6}\\\\\huge \purple{\boxed {x = 8}}\\m\angle ABC = (5x+11)°\\m\angle ABC = (5\times 8+11)°\\m\angle ABC = (40+11)°\\\huge \orange{\boxed {m\angle ABC = 51°}} \\$

7 0

2 years ago

HELP!!!!

ss7ja [257]

<h3>3 Answers: Choice D, Choice E, Choice F</h3>

============================================================

Explanation:

The inequality 6x - 10y ≥ 9 solves to y ≤ (3/5)x - 9/10 when you isolate y.

Graph the line y = (3/5)x - 9/10 and make this a solid line. The boundary line is solid due to the "or equal to" as part of the inequality sign. We shade below the boundary line because of the "less than" after we isolated for y.

Now graph all of the points given as I've done so in the diagram below. The points in the blue shaded region, or on the boundary line, are part of the solution set. Those points are D, E and F.

We can verify this algebraically. For instance, if we weren't sure point E was a solution or not, we would plug the coordinates into the inequality to get...

6x - 10y ≥ 9

6(5) - 10(2) ≥ 9 .... plug in (x,y) = (5,2)

30 - 20 ≥ 9

10 ≥ 9 ... this is a true statement

Since we end up with a true statement, this verifies point E is one of the solutions. I'll let you check points D and F.

-----------

I'll show an example of something that doesn't work. Let's pick on point A.

We'll plug in (x,y) = (-1,1)

6x - 10y ≥ 9

6(-1) - 10(1) ≥ 9

-6 - 10 ≥ 9

-16 ≥ 9

The last inequality is false because -16 is smaller than 9. So this shows point A is not a solution. Choices B and C are non-solutions for similar reasons.