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Bumek [7]
3 years ago
9

The floor of a canyon has an elevation of −14.5−14.5 feet. Erosion causes the elevation to change by −1.5−1.5 feet per year. How

many years will it take for the canyon floor to have an elevation of −31−31 feet?
Mathematics
1 answer:
Alona [7]3 years ago
4 0
<span>The canyon starts at an elevation of (-14.5). After each year, the elevation drops by (-1.5). The equation that could be written for this, then, would be (-14.5 - 1.5x = -31). First, we could add 14.5 to both sides of the equation to isolate the unknown. This would leave -1.5x = -16.5. Next, we can remove the negative signs so both sides of the equation are positive and easier to work with. This leaves 1.5x = 16.5. Finally, dividing both sides by 1.5 gives us "x = 11", which means that after 11 years, the canyon floor will be at -31 feet.</span>
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8 0
3 years ago
Read 2 more answers
Part 3 - Discussion/Explanation Question
SpyIntel [72]

Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

\frac{1}{x - 5}

Set the expression in the denominator equal to 0, because you can't divide by 0.

x - 5 = 0

x = 5

So the vertical asymptote is x=5.

Disclaimer if you see something like this

\frac{(x - 5)(x + 3)}{(x - 5)}

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

\frac{1}{x}

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

a \times  {x}^{0}

For example,

1 = 1 \times  {x}^{0}

2 =  2 \times {x}^{0}

And so on,

and

x =  {x}^{1}

So our equation is basically

\frac{1 \times  {x}^{0} }{ {x}^{1} }

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

\frac{3 {x}^{2} }{ {x}^{2}  + 1}

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

x =  \frac{3}{1}

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

8 0
2 years ago
If pq=3 and pq+rs=5, then 3+rs=5 is an example of the _________.
Mrrafil [7]
<span>This is an example of the substitution property of equality, meaning that if pq = 3, you may substitute 3 for pq in another related equation. So if pq+rs=5 is true, then 3+rs=5 is true as well.</span>
8 0
3 years ago
If f(x) = x2 - 1 and g(x) = 2x - 3, what is the domain of (fog)(x)?
ladessa [460]

Answer:

domain will be ( -∞, ∞)

Step-by-step explanation:

The given functions are f(x) = x² - 1 and g(x) = 2x - 3

We have to find domain of (fog)(x)

We will find the function (fg)(x) first.

(fog)(x) = f[g(x)]

         = (2x - 3)²

         = 4x² + 9 - 12x - 1

        = 4x² - 12x + 8

       = 4 (x² - 3x + 2)

The given function is defined for all values of x.

Therefore, domain will be ( -∞, ∞)

<u>brainly.com/question/2458431</u>

4 0
2 years ago
A teacher has been given £35 to spend on pencils and rulers.
SSSSS [86.1K]
Packet of 12 pencils = p
Packet of 30 rulers = r

p = £2.15
r = £6.00

£35 = 4r + xp
£35 = 4(£6.00) + x(£2.15)
£35 = £24 + £2.15x
£11 = £2.15x
x = 5.12

Since she can buy 5.12 packets of pencils, it is equal to 5 packets of pencils.
3 0
2 years ago
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