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3 years ago

8

What is the equation of a line that contains the point (2,-5) and is parallel to the line y=3x-4

1 answer:

aev [14]3 years ago

7 0

Answer:

$\large\boxed{y=3x-11}$

Step-by-step explanation:

$\text{Let}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\===================================\\\\y=3x-4\to m_1=3\\\\\text{Therefore}\ m_2=3.\\\\\text{We have the equation:}\ y=3x+b.\\\\\text{Put the coorsdinates of the point (2, -5) to the equation of the line}\\\\-5=3(2)+b\\-5=6+b\qquad\text{subtract 6 from both sides}\\-11=b\to b=-11$

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Answer:

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Step-by-step explanation:

We are asked to divide our given fraction: $\frac{45^{45}*15^{15}}{75^{30}}$ .

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Using product rule of exponents $(a*b)^n=a^n*b^n$ we can write:

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Substituting these values in our division problem we will get,

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Using power rule of exponents $a^m*a^n=a^{m+n}$ we will get,

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Using quotient rule of exponent $\frac{a^m}{a^n}=a^{m-n}$ we will get,

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$\frac{3^{45}*15^{30}}{5^{30}}$

Using product rule of exponents $(a*b)^n=a^n*b^n$ we will get,

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$\frac{3^{45}*3^{30}*5^{30}}{5^{30}}$

Upon canceling out $5^{30}$ we will get,

$3^{45}*3^{30}$

Using power rule of exponents $a^m*a^n=a^{m+n}$ we will get,

$3^{45+30}$

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Therefore, our resulting quotient will be $3^{75}$ .

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