I need help with these

Great Mountain Park: All day pass: 20.25 Mountain coaster: 4.50 Tilt a Loop: 4.75 Water coaster: 4.00 Fast drop: 3.50 Swing a Li

Black_prince [1.1K]

You would get £22.75 change from an £100 bill.

8 0

3 years ago

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How many sixths are in 3 and 1 third ??

lorasvet [3.4K]

1 whole thing = 6 sixths
3 whole things = 18 sixths

1 third of a whole thing = 2 sixths of it

3 wholes plus 1 third = 18 sixths plus 2 sixths = 20 sixths

6 0

3 years ago

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Explain how to determine if the number is a solution to the equation. n + 6 = 33 for n = 27

stepan [7]

If you subtract 6 from 33 youll end up with 27 and therefore 27 plus 6 equals 33

8 0

3 years ago

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If the sum of six consecutive even integers is 390 what is the smallest of the six integers

saveliy_v [14]

The answer is 60 but im going to explain it down below if you want to know how to do it :)

first because we dont know the values of the 6 numbers we have to lay it out like this.
x
x+2
x+4
x+6
x+8
x+10
-------- (then we add it together)
6x+30 = 390 (the sum of the 6 numbers)

using this equation we find out what x is

so we do 390-30 to find what 6x is, which is 360
because now we know that 6x=360, we divide 360 by 6 to find out what x is
360 divided by 6 =60..... x = 60

so now we know that the smallest integer is 60
to check the answer we could add the 6 numbers which are:
60+62+64+66+68+70=390 :)

so the smallest integer is 60

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=

BARSIC [14]

<h3>Answer: -2</h3>

======================================================

Work Shown:

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\$

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\$

-------------------

Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

$\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\$

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\$

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0

3 years ago