Simplify 4/24 to lowest terms.

Complete the square to transform the quadratic equation into the form

faltersainse [42]

The square of a binomial is written like

$(x\pm a)^2 = x^2\pm 2ax + a^2$

In your case, you have $2ax=-8x$ , which implies $a=-4$

So, we want to write

$(x-4)^2 = x^2-8x+16$

But our left hand side is

$x^2-8x-10$

If we add 26 to both sides, we have

$x^2 - 8x - 10 +26 = 18+26 \iff x^2-8x+16=44 \iff (x-4)^2=44$

7 0

3 years ago

Question 6. Geometry question.

kupik [55]

B. if two lines are perpendicular, then they intersect to form four right angles.

4 0

3 years ago

Barry got a samll sandwich for lunch and ate 1/2 of his sandwich. Jonah got a large sandwich and ate 1/2 of his sandwich. Did Jo

xeze [42]

No, they did not eat the same amount. Barry ate less because his sandwich was smaller, so there would be a smaller half. Jonah ate more because his sandwich was bigger, so he would have eaten a bigger half.

8 0

3 years ago

Please help I will give Brainliest please!

WITCHER [35]

Part (a)

The domain is the set of allowed x inputs of a function.

The graph shows that x = 0 is not allowed because of the vertical asymptote located here. It seems like any other x value is fine though.

<h3>Domain: set of all real numbers but $x \ne 0$ </h3>

To write this in interval notation, we can say $(-\infty, 0) \cup (0, \infty)$ which is the result of poking a hole at 0 on the real number line.

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The range deals with the y values. The graph makes it seem like it stretches on forever in both up and down directions. If this is the case, then the range is the set of all real numbers.

<h3>Range: Set of all real numbers</h3>

In interval notation, we would say $(-\infty, \infty)$ which is almost identical to the interval notation of the domain, except this time of course we aren't poking at hole at 0.

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Part (b)

<h3>The x intercepts are x = -4 and x = 4</h3>

We can compact that to the notation $x = \pm 4$

These are the locations where the blue hyperbolic curve crosses the x axis.

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Part (c)

<h3>Answer: There aren't any horizontal asymptotes in this graph.</h3>

Reason: The presence of an oblique asymptote cancels out any potential for a horizontal asymptote.

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Part (d)

The vertical asymptote is located at x = 0, so the equation of the vertical asymptote is naturally x = 0. Every point on the vertical dashed line has an x coordinate of zero. The y coordinate can be anything you want.

<h3>Answer: x = 0 is the vertical asymptote</h3>

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Part (e)

The oblique or slant asymptote is the diagonal dashed line.

It goes through (0,0) and (2,6)

The equation of the line through those points is y = 3x

If you were to zoom out on the graph (if possible), then you should notice the branches of the hyperbola stretch forever upward but they slowly should approach the "fencing" that is y = 3x. The same goes for the vertical asymptote as well of course.

<h3>Answer: Oblique asymptote is y = 3x</h3>

5 0

2 years ago

I need help with solving this

leva [86]

Answer:

49

Step-by-step explanation:

Positive 49 not -49

3 0

3 years ago