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TEA [102]
3 years ago
14

Why can a problem with extra information be difficult to solve

Mathematics
2 answers:
Rus_ich [418]3 years ago
8 0
That and your brain is trying hard to absorb info and you have to be capable of filtering what you need and don't need. You have to know what the math teacher is thinking and what formula he wants you to use. It's kind of unfair sometimes.
jok3333 [9.3K]3 years ago
5 0
You use the extra info and end up getting the wrong answer
You might be interested in
Find the intercepts and the vertical and horizontal asymptotes, and then use them to sketch a graph of the function. f(x)=x+2/x^
hoa [83]

We have the function:

f(x)=x+\frac{2}{x^2}-16.

We must find:

0. the intercepts,

,

1. the vertical and horizontal asymptotes.

1) x-intercepts

The x-intercepts are given by the x values such that f(x) = 0. So we must find the values of x that satisfies the equation:

f(x)=x+\frac{2}{x^2}-16=0.

Solving for x, we get:

\begin{gathered} x+\frac{2}{x^2}-16=0 \\ x\cdot x^2+2-16\cdot x^2,\text{ }x\ne0, \\ x^3-16x^2+2=0. \end{gathered}

The real roots of this equation are:

\begin{gathered} x_1\approx15.9922, \\ x_2\approx0.35757, \\ x_3\approx-0.34975. \end{gathered}

So the x-intercepts are the points:

\begin{gathered} P_1=(15.9922,0), \\ P_2=(0.35757,0), \\ P_3=(-0.34975,0)\text{.} \end{gathered}

2) y-intercepts

The y-intercepts are given by the y values such that x = 0. Replacing x = 0 in the definition f(x), we see that the denominator of the second term diverges. So we conclude that there are no y-intercepts.

3) Vertical asymptotes

Vertical asymptotes are vertical lines near which the function grows without bound. From point 2, we know that the function grows without limit when x goes to zero. So one vertical asymptote is:

x=0.

4) Horizontal asymptotes

Horizontal asymptotes are horizontal lines that the graph of the function approaches when x → ±∞. We consider the limit of the function f(x) when x → ±∞:

\lim _{x\rightarrow\pm\infty}f(x)=\lim _{x\rightarrow\pm\infty}(x+\frac{2}{x^2}-16)\rightarrow\pm\infty.

We see that the function does not tend to any constant value when x → ±∞. So we conclude that there are no horizontal asymptotes.

5) Oblique asymptotes

When a linear asymptote is not parallel to the x- or y-axis, it is called an oblique asymptote or slant asymptote.

A function ƒ(x) is asymptotic to the straight line y = mx + n (m ≠ 0) if

{\displaystyle\lim _{x\to+\infty}\mleft[f(x)-(mx+n)\mright]=0\, {\mbox{ or }}\lim _{x\to-\infty}\mleft[f(x)-(mx+n)\mright]=0.}

We consider the line given by:

y=mx+n=x-16.

We compute the limit:

\begin{gathered} \lim _{x\rightarrow\pm\infty}(f(x)-(x-16)) \\ =\lim _{x\rightarrow\pm\infty}((x+\frac{2}{x^2}-16)-(x-16)) \\ =\lim _{x\rightarrow\pm\infty}(\frac{2}{x^2}) \\ =0. \end{gathered}

So we have proven that f(x) has the oblique asymptote:

y=x-16.

6) Graph

Plotting the intercepts and the asymptotes, we get the following graph:

Answer

1) x-intercepts: (-0.34975, 0), (0.35757, 0), (15.9922, 0)

2) y-intercepts: none

3) Vertical asymptotes: x = 0

4) Horizontal asymptotes: none

5) Oblique asympsotes: y = x -16

6) Graph

6 0
1 year ago
Use substitution to solve the systems <br> y=5x<br> x+y=-6<br><br> x+2y=8<br> 3x-4y=4
VladimirAG [237]
Your answers are:
1.(-1,-5)
2.(4,2)
4 0
3 years ago
(Please show all work) 1. Find f(6) if f(x)= 1/2x + x - 4
dsp73

Answer:

\large\boxed{1.\ f(6)=5\ or\ f(6)=20}\\\boxed{2.\ d=5}\\\boxed{3.\ vertex=(-3,\ -1)}

Step-by-step explanation:

1.\\\text{Put x = 6 to the equation of a function}\ f(x):\\\\\text{If}\ f(x)=\dfrac{1}{2}x+x-4\to f(6)=\dfrac{1}{2}(6)+6-4=3+6-4=5\\\\\text{If}\ f(x)=\dfrac{1}{2}x^2+x-4\to f(6)=\dfrac{1}{2}(6^2)+6-4=\dfrac{1}{2}(36)+2=18+2=20\\\\2.\\\text{The formula of a distance between two points:}\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\\text{We have the points (2, -4) and (5, -8). Substitute:}\\\\d=\sqrt{(5-2)^2+(-8-(-4))^2}=\sqrt{3^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5

3.\\\text{The vertex formula of a parabola:}\\\\y=a(x-h)^2+k\\\\(h,\ k)-vertex\\\\\text{We have}\ y=(x+3)^2-1=(x-(-3))^2-1\\\\\text{Therefore the vertex is:}\ (-3,\ -1).

8 0
3 years ago
PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU B
ozzi

Answer:

72

Step-by-step explanation:

Just multiply all them

7 0
3 years ago
Helpppppppppppppppppppppppppppppppppppppppppp
solong [7]

Answer:

Please check explanations for answer

Step-by-step explanation:

Here, we want to check the given list and state the true values;

1. This is true

Compound interest has the exponent nt; simple interest has no exponent

2. This is false

Simple interest is only earned on the principal which is the amount deposited

3. True

Due to its multiplication chain, it earns more money as we can compound more than once within the given time frame

4. This is false; check explanations above

5. This is true

Its value is only based on the amount invested originally

6. This is false

simple interest earns the same amount annually based on rate

7. This is false

Both considers the time factor t in their formulas

8. This is true

The interest and principal forms the principal for another compounding time

5 0
3 years ago
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