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3 years ago

Write 4y−10=0 in slope-intercept form

Mathematics

2 answers:

Natalka [10]3 years ago

6 0

Answer:

y= 5/2

Step-by-step explanation:

4y-10=0

+10 +10 <-- add 10 to both sides

4y=10

4y/4=10/4 <-- divide by 4 to isolate y

y=10/4

y=5/2 <-- simplify

pogonyaev3 years ago

4 0

Answer:

y=5/2

Step-by-step explanation:

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3 years ago

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aivan3 [116]

Answer:

We verified that $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

Hence proved

Step-by-step explanation:

Given equation is $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

We have to prove that $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

That is to prove that LHS=RHS

Now taking RHS

$\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

$=\frac{a+b+c}{2}[a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2]$ (using $(a-b)^2=a^2-2ab+b^2$ )

$=\frac{a+b+c}{2}[2a^2-2ab+2b^2-2bc+2c^2-2ac]$ (adding the like terms)

$=\frac{a+b+c}{2}[2a^2+2b^2+2c^2-2ab-2bc-2ac]$

$=\frac{a+b+c}{2}\times 2[a^2+b^2+c^2-ab-bc-ac]$

$=a+b+c[a^2+b^2+c^2-ab-bc-ac]$

Now multiply the each term to another each term in the factor

$=a^3+ab^2+ac^2-a^2b-abc-a^2c+ba62+b^3+bc^2-ab^2-b^2c-abc+ca^2+cb^2+c^3-abc-bc^2-ac^2]$

$=a^3+b^3+c^3-3abc$ (adding the like terms and other terms getting cancelled)

$=a^3+b^3+c^3-3abc$ =LHS

Therefore LHS=RHS

Therefore $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

Hence proved.

8 0

4 years ago