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3 years ago

Identify an equation in slope-intercept from for the line parallel to y=-3x + 7 that passes through (2, -4)

Mathematics

1 answer:

alexandr1967 [171]3 years ago

7 0

The slope-intercept form:

y = mx + b

m - slope

b - y-intercept

The parallel lines have the same slope.

We have y = -3x + 7 → m = -3. Therefore y = -3x + b.

Put the coordinates of the point (2, -4) to the equation of a line:

$-4=-3(2)+b$

$-4=-6+b$ <em>add 6 to both sides</em>

$2=b\to b=2$

<h3>Answer: y = -3x + 2</h3>

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Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.

Lelechka [254]

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1) $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

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Step-by-step explanation:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}$

Solving using exponent rule: $a^{-m}=\frac{1}{a^m}$

$\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32$

So, $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4}$

Using the exponent rule: $a^m.a^n=a^{m+n}$

We have:

$2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}$

We also know that: $a^{-m}=\frac{1}{a^m}$

Using this rule:

$2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}$

So, $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2$

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$(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}$

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$\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32$

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