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2 years ago

PLEASE HELP I HAVE AN HOUR LEFT!!

Mathematics

1 answer:

Yuki888 [10]2 years ago

5 0

The statement that correctly describes the horizontal asymptote of g(x) is:

Limit of g (x) as x approaches plus-or-minus infinity = 6, so g(x) has an asymptote at y = 6.

<h3>What are the asymptotes of a function f(x)?</h3>

The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator.

The horizontal asymptote is the limit of f(x) as x goes to infinity, as long as this value is different of infinity.

In this problem, the function is:

$g(x) = \frac{42x^3 - 15}{7x^3 - 4x^2 - 3}$

The horizontal asymptote is given as follows:

$y = \lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} \frac{42x^3 - 15}{7x^3 - 4x^2 - 3} = \lim_{x \rightarrow \infty} \frac{42x^3}{7x^3} = \lim_{x \rightarrow \infty} 6 = 6$

Hence the correct statement is:

Limit of g (x) as x approaches plus-or-minus infinity = 6, so g(x) has an asymptote at y = 6.

More can be learned about asymptotes at brainly.com/question/16948935

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The equation for the pH of a substance is pH = –log[H+], where H+ is the concentration of hydrogen ions. Which equation models t

Zanzabum

Answer: [H+] = 10^-7.2

Step-by-step explanation:

Given that the PH of the solution = 7.2

Using the formula pH = –log[H+]. To get the H+ concentration from the pH, raise both sides by the base of 10. Then we have

10^ -pH = H+. with pH of 7.2,

Thus the answer to this problem is

[H+] = 10^-7.2

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3 years ago

A napkin is folded into an isosceles triangle, triangle ABC, and placed on a plate, as shown. The napkin has a perimeter of 38 c

gavmur [86]

The area of plate which is covered by the napkin is equal to the area of folded napkin.

Before finding the area of folded napkin, we should find the dimension of the napkin.
Find the length of the leg side, the two leg has similar length.
perimeter = 38
l + l + 8 = 38
2l + 8 = 38
2l = 30
l = 15
Each leg is 15 cm long.

Find the area of folded napkin
area = 1/2 × s × s × sin of angle between the sides
area = 1/2 × 15 × 15 × sin 30°
area = 1/2 × 15 × 15 × 1/2
area = 225/4
area = 56.25
to the nearest whole number >> 56 cm²
The answer is third option

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3 years ago

Read 2 more answers

Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat

eimsori [14]

Answer:

There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term $2x^2$ , which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.

2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

We recall that the x-position of the vertex for a quadratic function of the form $f(x)=ax^2+bx+c$ is given by the expression:

$x_v=\frac{-b}{2a}$

Since in our case $a=2$ and $b=-3$ , we get that the x-position of the vertex is:

$x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

$y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}$

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

$f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6$