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2 years ago

What is the equation of the line that passes through the point (-2,7) and has a slope of zero

Mathematics

1 answer:

Lina20 [59]2 years ago

5 0

Answer:

Equation of the line that passes through the point (-2,7) and has a slope of zero is given as

$y=7$

Step-by-step explanation:

Equation of line is given as

$y-y_1=m(x-x_1)$ -----------(1)

Here

$m=0 \ \ \ and \ \ \ (x_1, y_1) = (-2, 7)$

Substituting values in equation (1)

$y-7=0(x+2)\\y=7$

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Which shows the following expression using positive exponents?

Hitman42 [59]

Answer:

1. option B is correct i.e., $\frac{-3a^{7}b^3b^1}{15a^{2}}$

2. option D is correct i.e., $\frac{a^3b^4}{ab^2}$ .

3. option B is correct i.e., $\frac{n^{10}}{16m^{6}}$ .

Step-by-step explanation:

1. Since the given equation is:

$\frac{-3a^{-2}b^3}{15a^{-7}b^{-1}}$

As we know that $a^{-x}=\frac{1}{a^x}$

So, the given equation can be represented as:

$\frac{-3a^{7}b^3b^1}{15a^{2}}$

2. Given equation is:

$\frac{a^3b^{-2}}{ab^{-4}}$

As we know that $a^{-x}=\frac{1}{a^x}$

so the given equation can be represented as:

$\frac{a^3b^4}{ab^2}$ .

3. Given equation is:

$(\frac{4mn}{m^{-2}n^6})^{-2}$

by using the properties $a^{-x}=\frac{1}{a^x}$ ,

and $a^x\times a^y=a^{x+y}$

$(\frac{4mn}{m^{-2}n^6})^{-2}$

= $(\frac{4n^1n^{-6}}{m^{-2}m^{-1}})^{-2}$

= $(\frac{4n^{1+(-6)}}{m^{-2+(-1)}})^{-2}$

= $(\frac{4n^{-5}}{m^{-3}})^{-2}$

= $\frac{4^{-2}n^{-5\times (-2)}}{m^{-3\times (-2)}}$

= $\frac{n^{10}}{16m^{6}}$ .

which is option B.

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What is the equation of the line that passes through the point (-2,7) and has a slope of zero​

What is the equation of the line that passes through the point (-2,7) and has a slope of zero