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const2013
7 days ago
10

 how do i slove  5/9v+w=z, for v

Mathematics
1 answer:
Leona [9.2K]7 days ago
3 0


We want v alone, so subtract w from both sides to move it over

This leaves 5/9 v = z - w

Next, eliminate the fraction multiplying v by multiplying both sides by its reciprocal

The reciprocal of 5/9 is 9/5

(5/9) × (9/5) cancel each other, leaving v isolated on the left

The right-hand side becomes 9/5 (z - w)

Combine to get:

v = 9/5 (z - w)

There you go!:D

 
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Which are the solutions of x2 = –5x + 8? StartFraction negative 5 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFract
tester [8842]

Answer:

x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}

Detailed solution:

Given:

The problem to solve is:

x^2=-5x+8

Convert the equation into the standard quadratic form ax^2+bx +c =0, where a,\ b,\ and\ c represent constants.

So, by adding 5x-8 to both sides, we get:

x^2+5x-8=0

Note that a=1,b=5,c=-8.

The roots of this quadratic are found by applying the quadratic formula given as:

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Substitute a=1,b=5,c=-8 into the formula and calculate for x.

x=\frac{-5\pm \sqrt{5^2-4(1)(-8)}}{2(1)}\\x=\frac{-5\pm \sqrt{25+32}}{2}\\x=\frac{-5\pm \sqrt{57}}{2}\\\\\\\therefore x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}

Hence, the roots are:

x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}

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