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lubasha [3.4K]
3 years ago
6

Multiply and simplify.

Mathematics
1 answer:
Igoryamba3 years ago
4 0
(2x-1)(6x-7)=2x\cdot 6x-2x\cdot7-1\cdot 6x+1\cdot7=\\\\=12x^2-14x-6x+7=\boxed{12x^2-20x+7}

Answer D.
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If HJ=2x+5, JK=3x-7, and KH=18 what is the value of x
weqwewe [10]

Answer:

  • x = 4

Step-by-step explanation:

<u>Given</u>

  • HJ=2x+5
  • JK=3x-7
  • KH=18

<u>We see J is between H and K, so </u>

  • KH = HJ + JK
  • 2x + 5 + 3x - 7 = 18
  • 5x - 2 = 18
  • 5x = 20
  • x = 4

3 0
2 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.

--------------

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.

=======================================================

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.

=======================================================

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.

=======================================================

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>

=======================================================

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
What’s the answer???
enyata [817]
Nikkis charges are 20,40,50,80. Dave’s charges are 25 and 37.5.
5 0
3 years ago
Nolan is using substitution to determine if 23 is a solution to the equation. Complete the statements.
vlada-n [284]

Answer: first, Nolan must substitute 23 for J To simplify, Nolan must subtract 16 form 23. 23 is a solution of the equation.

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Solve for x. <br> 35x = 21
emmasim [6.3K]
X = 21/35

x = 3/5 

Explanation: divide both sides by 35 and reduce the fraction 21/35 to get 3/5.

7 0
3 years ago
Read 2 more answers
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