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Iteru [2.4K]
3 years ago
9

Plz help me with this

Mathematics
1 answer:
Lady_Fox [76]3 years ago
5 0
Basically it’s asking when is the function g(x) greater than the function of f(x), which are both graphed on the graph. g(x) needs to be above f(x) for it to be greater, and that is shows between 0-2 and 4+, making the answer A
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Add 2/5 plus 27/9 -0
Leviafan [203]
2/5 plus 27/9 minus 0 equals 0
4 0
3 years ago
What measurement that is equal to 20 liters
Aliun [14]

Answer:5.28344 gallons

Step-by-step explanation: idk i just looked it up lol.

8 0
3 years ago
Read 2 more answers
What is the equation of the oblique asymptote?<br> h(x) = x² – 3x - 4/x + 2
NNADVOKAT [17]

Simplifying h(x) gives

h(x) = (x² - 3x - 4) / (x + 2)

h(x) = ((x² + 4x + 4) - 4x - 4 - 3x - 4) / (x + 2)

h(x) = ((x + 2)² - 7x - 8) / (x + 2)

h(x) = ((x + 2)² - 7 (x + 2) - 14 - 8) / (x + 2)

h(x) = ((x + 2)² - 7 (x + 2) - 22) / (x + 2)

h(x) = (x + 2) - 7 - 22/(x + 2)

h(x) = x - 5 - 22/(x + 2)

An oblique asymptote of h(x) is a linear function p(x) = ax + b such that

\displaystyle \lim_{x\to\pm\infty} h(x) - p(x) = 0

In the simplified form of h(x), taking the limit as x gets arbitrarily large, we obviously have -22/(x + 2) converging to 0, while x - 5 approaches either +∞ or -∞. If we let p(x) = x - 5, however, we do have h(x) - p(x) approaching 0. So the oblique asymptote is the line y = x - 5.

4 0
2 years ago
What is the range of the relation[(-5,0),(-4,-3),(-3,4),(-1,0),(-4,-1)]
vesna_86 [32]

The range is the set of values that the relation takes for the domain for which it is defined.

Is the set of values of the dependent variable.

In this case, we have the relation [(-5,0),(-4,-3),(-3,4),(-1,0),(-4,-1)]​.

The dependant variable takes the values: 0, -3, 4, 0 and -1. Some values are repeated.

We put the repeated values only once and sort to write the range.

Then, the range is R = {-3, -1, 0, 4}.

Answer: Range = {-3, -1, 0, 4}

4 0
11 months ago
Need help<br> Question in the screenshot below<br> NO spam<br> Please show your work
Alja [10]
<h3>Answer:   x = 6</h3>

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

Work Shown:

\log_{4}(x+10)+\log_{4}(x-2)=\log_{4}(64)\\\\\log_{4}\left((x+10)(x-2)\right)=\log_{4}(64)\\\\(x+10)(x-2)=64\\\\x^2-2x+10x-20=64\\\\x^2-2x+10x-20-64=0\\\\

x^2+8x-84=0\\\\(x+14)(x-6)=0\\\\x+14=0 \ \text{ or } \ x-6=0\\\\x=-14 \ \text{ or } \ x=6\\\\

Those are the possible solutions, but plugging x = -14 back into the original equation will lead to an error. So we rule x = -14 out

x = 6 works as a solution however

3 0
2 years ago
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