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3 years ago

What is the equation of the line that passes through the point (-6, 4) and has a slope of -5/6

Mathematics

1 answer:

tangare [24]3 years ago

5 0

Answer:

$y = -\frac{5}{6}x -1$

Step-by-step explanation:

So, we know the slope is -5/6. From here we can write this equation:

$y = -\frac{5}{6}x + b$

In the question they give us both the x and y in this equation (-6,4)

Plug them in to the equation and solve for b:

$4 = -\frac{5}{6} (-6) + b\\4 = \frac{30}{6} + b\\4 = 5 + b \\-1 = b$

Now, we can write the equation as follows:

$y = -\frac{5}{6}x -1$

<em>I hope this helps!!</em>

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We can conclude that the radius of our spherical balloon is approximately 3 inches long. Therefore, the correct answer is A. 3 in.

2. Lets simplify each one of the expressions first:
F. $\sqrt{x^2}$ since the radicand is elevated to the same number as the index of the radical, we can cancel the radical and the exponent:
$\sqrt{x^2} =x$
Since $x\ \textgreater \ 0$ , this expression is equal to $x$ .
G. $\frac{1}{2} \sqrt[3]{8x^3}$ 8 can be expressed as $8=2*2*2=2^3$ , so we can rewrite our radicand:
$\frac{1}{2} \sqrt[3]{8x^3} =\frac{1}{2} \sqrt[3]{2^3x^3} = \frac{1}{2} \sqrt[3]{(2x)^3}$
Since the radicand is elevated to the same number as the index of the radical, we can cancel the radical and the exponent:
$\frac{1}{2} \sqrt[3]{(2x)^3}= \frac{1}{2} (2x)$
Now, we can cancel the 2 in the denominator with the one in the numerator:
$\frac{1}{2} (2x)=x$
Since $x\ \textgreater \ 0$ , this expression is equal to $x$ .
H. $\sqrt[3]{-x^3}$ The radicand is elevated to the same number as the index of the radical, so we can cancel the radical and the exponent:
$\sqrt[3]{-x^3}=-x$
Since $x\ \textgreater \ 0$ , this expression is NOT equal to $x$ .

We can conclude that the correct answer is H. $\sqrt[3]{-x^3}$ .

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The correct answer is D. no real root found.

4. This time we are going to take a different approach. We are going to simplify $\sqrt{a^2(x+a^2)}$ first, and then, we are going to compare the result with our given options:
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$\sqrt{4x^2y^4} = \sqrt{4} \sqrt{x^2} \sqrt{y^4}$
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