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pantera1 [17]
3 years ago
9

Sun spot activity goes in an eleven-year cycle. The last cycle started in 2008 and ended in 2018. Here is the activity per year

Mathematics
1 answer:
KIM [24]3 years ago
3 0

Answer:

19999999999

Step-by-step explanation:

xdnsedfeadwh oASDSADAS

You might be interested in
A) Evaluate the limit using the appropriate properties of limits. (If an answer does not exist, enter DNE.)
Gelneren [198K]

For purely rational functions, the general strategy is to compare the degrees of the numerator and denominator.

A)

\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \boxed{\frac27}

because both numerator and denominator have the same degree (2), so their end behaviors are similar enough that the ratio of their coefficients determine the limit at infinity.

More precisely, we can divide through the expression uniformly by <em>x</em> ²,

\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \lim_{x\to\infty} \frac{2-\dfrac5{x^2}}{7+\dfrac1x-\dfrac3{x^2}}

Then each remaining rational term converges to 0 as <em>x</em> gets arbitrarily large, leaving 2 in the numerator and 7 in the denominator.

B) By the same reasoning,

\displaystyle \lim_{x\to\infty} \frac{5x-3}{2x+1} = \boxed{\frac52}

C) This time, the degree of the denominator exceeds the degree of the numerator, so it grows faster than <em>x</em> - 1. Dividing a number by a larger number makes for a smaller number. This means the limit will be 0:

\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \boxed{0}

More precisely,

\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \lim_{x\to-\infty}\frac{\dfrac1x-\dfrac1{x^2}}{1+\dfrac8{x^2}} = \dfrac01 = 0

D) Looks like this limit should read

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2}

which is just another case of (A) and (B); the limit would be

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = -1

That is,

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = \lim_{t\to\infty}\frac{\dfrac1{t^{3/2}}+1}{\dfrac3t-1} = \dfrac1{-1} = -1

However, in case you meant something else, such as

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t+t^2}}{3t-t^2}

then the limit would be different:

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t^2}\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{t\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{\sqrt{\dfrac1t+1}}{3-t} = 0

since the degree of the denominator is larger.

One important detail glossed over here is that

\sqrt{t^2} = |t|

for all real <em>t</em>. But since <em>t</em> is approaching *positive* infinity, we have <em>t</em> > 0, for which |<em>t</em> | = <em>t</em>.

E) Similar to (D) - bear in mind this has the same ambiguity I mentioned above, but in this case the limit's value is unaffected -

\displaystyle \lim_{x\to\infty} \frac{x^4}{\sqrt{x^8+9}} = \lim_{x\to\infty}\frac{x^4}{\sqrt{x^8}\sqrt{1+\dfrac9{x^8}}} = \lim_{x\to\infty}\frac{x^4}{x^4\sqrt{1+\dfrac9{x^8}}} = \lim_{x\to\infty}\frac1{\sqrt{1+\dfrac9{x^8}}} = \boxed{1}

Again,

\sqrt{x^8} = |x^4|

but <em>x</em> ⁴ is non-negative for real <em>x</em>.

F) Also somewhat ambiguous:

\displaystyle \lim_{x\to\infty}\frac{\sqrt{x+5x^2}}{3x-1} = \lim_{x\to\infty}\frac{\sqrt{x^2}\sqrt{\dfrac1x+5}}{3x-1} = \lim_{x\to\infty}\frac{x\sqrt{\dfrac1x+5}}{3x-1} = \lim_{x\to\infty}\frac{\sqrt{\dfrac1x+5}}{3-\dfrac1x} = \dfrac{\sqrt5}3

or

\displaystyle \lim_{x\to\infty}\frac{\sqrt{x}+5x^2}{3x-1} = \lim_{x\to\infty}x \cdot \lim_{x\to\infty}\frac{\dfrac1{\sqrt x}+5x}{3x-1} = \lim_{x\to\infty}x \cdot \lim_{x\to\infty}\frac{\dfrac1{x^{3/2}}+5}{3-\dfrac1x} = \frac53\lim_{x\to\infty}x = \infty

G) For a regular polynomial (unless you left out a denominator), the leading term determines the end behavior. In other words, for large <em>x</em>, <em>x</em> ⁴ is much larger than <em>x</em> ², so effectively

\displaystyle \lim_{x\to\infty}(x^4-2x) = \lim_{x\to\infty}x^4 = \boxed{\infty}

6 0
3 years ago
BRAINLIEST FOR RIGHT ANSWERS!
exis [7]

Answer:

The formula for finding volume of a cone is (in this case),

V=(3.14)(r^2)(h/3)

With this formula in mind, you would simply substitute in.

V=(3.14)(2^2)(9/3)

You could work that out for yourself on your own, but to save you the time, it is 37.68 in. cubed.

Step-by-step explanation:

hope i helped

6 0
3 years ago
Read 2 more answers
Sierra has plotted two vertices of a rectangle at (3,2) and (8,2). What is the length of the sides of the rectangle
Eddi Din [679]

Answer:

5units

Step-by-step explanation:

Given the two vertices of a rectangle at (3,2) and (8,2). We are to calculate the distance between the two points.

Using the formula

D = √(x2-x1)²+(y2-y1)²

D = √(2-2)²+(8-3)²

D = √0+5²

D =  √25

D = 5

Hence the length of the side of the rectangle is 5units

4 0
3 years ago
What the equivalent fraction 1/3
emmasim [6.3K]
2/6, 3/9, 4/12, 5/15, 6/18, 7/21, 8/24 etc.
7 0
4 years ago
Read 2 more answers
You get a 15% raise at your job and now make $13.50 per hour. What was you previous rate of pay per hour?
Anestetic [448]
<h2>We know:</h2>

You currently make $13.50 an hour. You've earned a 15% pay raise. You want to find out your previous rate of pay per hour.


<h2>Steps:</h2>

1.) Convert 15% to a decimal, which equals 0.15


2.) Multiply your current earnings with the percentage in decimal form (0.15)

$13.50 (current pay per hour) x 0.15 (percentage in decimal form) = $2.025 (difference between your previous wage and your current wage)


3.) Subtract the difference of your previous wage and current wage with your current earnings

$13.50 (current pay per hour) - $2.025 (difference between previous and current wage) = $11.475 (previous pay per hour)


4.) Since this is talking about wage, rounding is not applicable. In other cases, your number would be rounded off to $11.48, however you can't add what you don't have, so you would have to round down.


<h2>You were previously making $11.47 an hour.</h2>
7 0
3 years ago
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