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Darya [45]
3 years ago
5

What was the most important lesson you have learned this year in math class? why is it important?

Mathematics
2 answers:
Elden [556K]3 years ago
8 0

Answer:

Step-by-step explanation:

Practical Lessons

Whenever I asked my fourth-grade teacher how to spell a word, she would tell me to look it up in the dictionary. Not only do I still automatically look up a word when I'm stumped on how to spell it but I also pick up the dictionary just to learn new ones.  

Bella Frank  

Bethlehem, New Hampshire

I learned to "free-write" in college, which means spending 10 minutes writing whatever is in your head. Now I free-write whenever I feel overwhelmed. I can organize my thoughts or jot down a to-do list. If I declutter my brain, I am more productive.

Denise La Voire

Corona Del Mar, California

Learning how to read opened doors for me. Reading seems to be passive―it's quiet and still―but its impact is anything but. As a child, I read to escape. As an adult, I teach reading, in English-as-a-second-language classes.

Jenna Rindo

Pickett, Wisconsin

As a college-bound high school senior, I took a typing class. Two college degrees later, I can tell you that learning how to type was the most useful skill I picked up in school.

Claudia Pratt

San Francisco, California

How to write a letter. Each week throughout second grade, we would write a letter to a fellow student in our class. This taught me how to use words to express myself and showed me how much joy there is in receiving a letter from a friend.

Stacey Kawamoto

Kapolei, Hawaii

Once I learned how to do research, I always knew how to go about finding the information I needed.

S. Metz

Piscataway, New Jersey

labwork [276]3 years ago
7 0
I think equations because you have to know how to use numbers
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Mr. Jimerson records the amount of snowfall that the local area recieves. 12am-1am: 0.3 inches | 1am-2am: 0.8 inches | 2am-3am:
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73.

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a)

\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3

b)

Since we can't divide by zero, we need to find when:

x^4-2x^2+144=0

But before, we can factor the numerator and the denominator:

\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}

Now, we can conclude that the vertical asymptotes are located at:

\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}

so, for x = -3:

\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty

For x = 4:

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