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3 years ago

3. Write the equation of a line that passes through the point (24, 25) and is

Mathematics

1 answer:

IceJOKER [234]3 years ago

4 0

Answer:

y=-1/3x + 33

Step-by-step explanation:

You can start by writing this in point slope form and converting to slope intercept later. Since the slope of the perpendicular line is y=3x-30, this line must have a slope of -1/3. It's point slope form is therefore:

y-25=-1/3(x-24)

Now, you can convert to slope intercept by isolating y:

y=-1/3(x-24)+25

y=-1/3x+8+25

y=-1/3x+33

Hope this helps!

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Which equation describes a line of symmetry for the figure shown? x = 0 y=−x y = x y = 0

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Answer:

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Step-by-step explanation:

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3 years ago

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Anon25 [30]

Answer:

f(x) = 3x

Step-by-step explanation:

The equation of a line passing through the origin is

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3 years ago

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2 years ago

Read 2 more answers

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nordsb [41]

Answer:

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Step-by-step explanation:

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2 years ago

4(3x^2y^4)^3 / (2x^3y^5)^4

tatyana61 [14]

Solving the expressions $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$ we get $\frac{27}{4x^{6}y^{8}}$

Step-by-step explanation:

We need to Solve the expression: $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

Solving:

$\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

$=\frac{4(3^3x^6y^{12})}{(2^4x^{12}y^{20})}\\=\frac{4(27x^6y^{12})}{(16x^{12}y^{20})}\\=\frac{108x^6y^{12}}{16x^{12}y^{20}}$

Applying exponent rule: $\frac{x^a}{x^b}=x^{a-b}$

$=\frac{108x^{6-12}y^{12-20}}{16}\\=\frac{27x^{-6}y^{-8}}{4}$

Another exponent rule says: $x^{-a}=\frac{1}{x^a}=$

$=\frac{27}{4x^{6}y^{8}}$

So, solving the expressions $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$ we get $\frac{27}{4x^{6}y^{8}}$

Keywords: Solving Exponents

Learn more about Solving Exponents at:

#learnwithBrainly

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2 years ago