All categories

3 years ago

Wx+yz=bc solve for x May someone help

Mathematics

2 answers:

Blababa [14]3 years ago

4 0

Wx+yz=bc
-yz. -yz
Wx=bc-yz
Wx/W=bc-yz/W
x=bc-yz/W

Natasha2012 [34]3 years ago

4 0

First isolate yz from wx, by subtracting it from both sides.

wx+yz=bc
wx+yz-yz=bc-yz

Now you should have: wx=bc-yz

From here, you isolate the w from wx in order for x to stand by itself:

wx/w=(bc-yz)/w

You might be interested in

Solve for x −9x+5≤17 OR 13x+25≤−1

arsen [322]

Answer:

(-∞ 2] ∪ [-4/3 ∞ )

Step-by-step explanation:

Given compound inequality,

$-9x+5\leq 17\text{ or }13x+25\leq -1$

$-9x\leq 12\text{ or }13x\leq -26$ ( Subtraction property of inequality )

$-x\leq \frac{12}{9}\text{ or }x\leq -\frac{26}{13}$ ( Division property of inequality )

$-x\leq \frac{4}{3}\text{ or }x\leq -2$

$x\geq -\frac{4}{3}\text{ or }x\leq -2$ ( a < b ⇒ - a > - b )

Hence, the solution of the given inequality is,

(-∞ -2] ∪ [-4/3 ∞ )

3 0

3 years ago

Read 2 more answers

A regular hexagon is dilated by a scale factor of 7/5 to create a new hexagon. How does the perimeter of the new hexagon compare

Airida [17]

<span>It's the same ratio, 7/5.</span>

3 0

3 years ago

Read 2 more answers

Which equation represents the equation of the parabola with focus (-3 3) and directrix y=7?

Artemon [7]

Answer:

The equation $y=\frac{-x^2-6x+31}{8}$ represents the equation of the parabola with focus (-3, 3) and directrix y = 7.

Step-by-step explanation:

To find the equation of the parabola with focus (-3, 3) and directrix y = 7. We start by assuming a general point on the parabola (x, y).

Using the distance formula $d = \sqrt {\left( {x_1 - x_2 } \right)^2 + \left( {y_1 - y_2 } \right)^2 }$ , we find that the distance between (x, y) is

$\sqrt{(x+3)^2+(y-3)^2}$

and the distance between (x, y) and the directrix y = 7 is

$\sqrt{(y-7)^2}$ .

On the parabola, these distances are equal so, we solve for y:

$\sqrt{(x+3)^2+(y-3)^2}=\sqrt{(y-7)^2}\\\\\left(\sqrt{\left(x+3\right)^2+\left(y-3\right)^2}\right)^2=\left(\sqrt{\left(y-7\right)^2}\right)^2\\\\x^2+6x+y^2+18-6y=\left(y-7\right)^2\\\\x^2+6x+y^2+18-6y=y^2-14y+49\\\\y=\frac{-x^2-6x+31}{8}$