4 years ago

I'm not sure how to answers these questions

Mathematics

1 answer:

steposvetlana [31]4 years ago

4 0

1.correct

2.correct

3.correct but put 37.5% instead

4.correct

5.correct but again 37.5

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Simplify the following : x^2y^-3 / x^5y^2

Likurg_2 [28]

Answer:

Step-by-step explanation:

$\frac{a^{m}}{a^{n}}=a^{m-n}\\\\a^{-m}=\frac{1}{a^{m}}\\\\\\\frac{x^{2}y^{-3}}{x^{5}y^{2}}=x^{(2-5)}*y^{(-3-2)}\\\\ = x^{-3}*y^{-5}\\\\=\frac{1}{x^{3}y^{5}}$

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3 years ago

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Determine the equation of a line that passes through A(2,5) and is parallel to the line defined by 3x−y+12=0. State the equation

Ksju [112]

Answer:

y = 3x - 1

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

3x - y + 12 = 0 ( subtract 3x + 12 from both sides )

- y = - 3x - 12 ( multiply through by - 1 )

y = 3x + 12 ← in slope- intercept form

with slope m = 3

Parallel lines have equal slopes , thus

y = 3x + c ← is the partial equation

To find c substitute (2, 5) into the partial equation

5 = 6 + c ⇒ c = 5 - 6 = - 1

y = 3x - 1 ← equation of parallel line

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3 years ago

Some roads in Poland have a speed limit of 140 km/h. How fast is this in miles per minute?

zheka24 [161]

1 .47 miles per minute

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How do i check an equation for extraneous solutions?

Marina86 [1]

Example:

$\sqrt x(x+1)=0$

This suggests two solutions, $x=0$ and $x=-1$ .

However, upon plugging these solutions back into the equation, you get

$\sqrt0(0+1)=0(1)=0$

which checks out, but

$\sqrt{-1}(1-1)=0$

does not because $\sqrt x$ is defined only for $x\ge0$ (assuming you're looking for real solutions only). So, we call $x=-1$ an extraneous solution, and the complete solution set (over the real numbers) is $x=0$ .

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KonstantinChe [14]

the answer is b for sure

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