Pleasehelp me with this im having trouble

If a,b,c are all non-zero numbers and a+b+c=0, prove that

Butoxors [25]

Given - a+b+c = 0

To prove that- 

a²/bc + b²/ac + c²/ab = 3

Now we know that

when x+y+z = 0,

then x³+y³+z³ = 3xyz

that means

 (x³+y³+z³)/xyz = 3 ---- eq 1)

Lets solve for LHS

LHS = a²/bc + b²/ac + c²/ab

we can write it as LHS = a³/abc + b³/abc + c³/abc

by multiplying missing denominators,

now take common abc from denominator and you'll get,

LHS = (a³+b³+c³)/abc --- eq (2)

Comparing one and two we can say that

(a³+b³+c³)/abc = 3

Hence proved,

a²/bc + b²/ac + c²/ab = 3

6 0

2 years ago

Randy has 53 inches of ribbon that he cuts into 8 pieces of equal length. What is the length of each piece of ribbon, in inches?

maw [93]

Answer:

6.625 inches or 6 5/8 inches

and the drop down hopefully will let you choose "between 6 and 7 inches"

Step-by-step explanation:

53/8 = 6.625

0.625 = 5/8

Let me know if you have questions about this. Have a good night.

6 0

2 years ago

The figure shows △ABC . BD is the angle bisector of ∠ABC .

Misha Larkins [42]

From angle bisector theorem, we know that:

$\dfrac{|AD|}{|DC|}=\dfrac{|BA|}{|BC|}\\\\\\\dfrac{|AD|}{|DC|}=\dfrac{8}{10}=\dfrac{4}{5}$

Moreover:

$|AD|+|DC|=6$

so:

$|DC|=6-|AD|$

and

$\dfrac{|AD|}{|DC|}=\dfrac{4}{5}\\\\\\\dfrac{|AD|}{6-|AD|}=\dfrac{4}{5}\qquad\qquad\text{cross multiplying}\\\\\\ 5\cdot|AD|=4\cdot\big(6-|AD|\big)\\\\5\cdot|AD|=24-4\cdot|AD|\\\\5\cdot|AD|+4\cdot|AD|=24\\\\9\cdot|AD|=24\qquad|:9\\\\\\|AD|=\dfrac{24}{9}=\boxed{\dfrac{8}{3}}$

3 0

3 years ago

Read 2 more answers

What is the value of x in the equation (StartFraction one-half EndFractionx + 12) = StartFraction one-half EndFraction(StartFrac

anastassius [24]

Answer:

$x =-24$