Determine the standard form of the equation of the line that passes through (-7,8) and (0,2)

Mathematics

1 answer:

Gre4nikov [31]3 years ago

5 0

Answer:

Step-by-step explanation:

y=mx+b

8=-7m+b

2=0m+b

subtract

6=-7m

-6/7=m

2=0m+b

2=b

y=-6/7x+2

change to standard from

y+6/7x=2

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6 ( f + 4 ) + 2 ( f + 4 )

Helga [31]

6(f+4) + 2(f+4)

Use the distributive property:

6f+24 + 2f + 8

Now combine like terms:

8f +32

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4 0

3 years ago

Read 2 more answers

Number 2 I need help on how to solve it

Hunter-Best [27]

Let's simplify the following expression:

$\text{ }\frac{\text{ \lparen2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{ }\frac{\text{ 6x}^5\text{y}^{-4}\text{z}^{-5}}{\text{ \lparen2x}^6\text{y}^3\text{z}^{-3})^2}$

We get,

$\text{ }\frac{(\text{2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{ }\frac{6x^5y^{-4}z^{-5}}{(2x^6y^3z^{-3})^2}$ $\text{ }\frac{(2x^4-y^6z^{-4})^2(6x^5y^{-4}z^{-5})}{(2x^2z^3)(2x^6y^3z^{-3})^2}$ $\text{ }\frac{(4x^8\text{ - 4x}^4y^6z^{-4}\text{ + y}^{12}z^{-8})(6x^5y^{-4}z^{-5})}{(2x^2z^3)(4x^{12}y^6z^{-6})}$ $\text{ }\frac{24x^{13}y^{-4}z^{-5}\text{ - 24x}^9y^2z^{-9}\text{ + 6x}^5y^8z^{-13}}{8x^{14}y^6z^{-3}}$