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If there are 30 problems and you solve 2 each minute, and 30/2 = 15, then it takes 15 minutes to finish the math homework!

First we add the numbers: 10,157 + 9987 = 20,144

To round, you take the hundreds number (in this case 1) and if it's 1,2,3,or 4, you round down, making the number just 20,000. If the hundreds number is 5,6,7,8, or 9, you round up, making the number 21,000. In this case, since the hundreds number is 1, we round down to 20,000.

Hope this helps!

(1-sqrt3)/(1+sqrt3) → (1-sqrt3)(1-sqrt3)/( (1+sqrt3)(1-sqrt3) ) → (1-2sqrt3+3)/(1-3)

→ (4 - 2sqrt3)/(-2) → -2 + sqrt3

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Fundamental theorem of calculus

\(\begin{array}{|rcll|} \hline g(s) &=& \displaystyle\int\limits_{8}^{s} f(t)\ dt =\displaystyle \int\limits_{8}^{s} (t-t^4)^5\ dt \\ &=& F(s) - F(8) \\\\ g'(s) &=& \dfrac{d}{ds} \displaystyle\int\limits_{8}^{s} f(t)\ dt \\\\ &=& \dfrac{d\ F(s)}{ds} - \dfrac{d\ F(8)}{ds} \quad &| \quad \dfrac{d\ F(8)}{ds} = 0, \text{ because } F(8) \text{ is constant.} \\\\ &=& \dfrac{d\ F(s)}{ds} \\\\ &=& f(s) \quad & |\quad f(t)=(t-t^4)^5 \\\\ \mathbf{g'(s)}& \mathbf{=} & \mathbf{(s-s^4)^5} \\ \hline \end{array}\)

The Formula is:

\(\begin{array}{|rcll|} \hline \dfrac{d}{ds} \displaystyle\int\limits_{a}^{s} f(t)\ dt = f(s) \\ \hline \end{array}\)

Are you asking how many ways can 5 people be scheduled with 1 per day for 5 days ?

there are 5 choices for the first day, 5 choices for hte second day etc  

=5^5 ways  =   5^5 =     3125 ways

You have succeeded!

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Imagine a right angled triangle with opposite = 2 and hypotenuse = x

Its adjacent is then sqrt(x^2 - 2^2), so the tangent of the angle is opposite/adjacent =  2/sqrt(x^2 - 2^2)

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I'll assume v is the final velocity, and that the acceleration is constant, in which case you can use:

(i) d = u*t +(1/2)*a*t^2   where d = distance,  u = initial velocity, a = acceleration and t = time

and

(ii)  v = u + a*t  

I'll also assume the acceleration is 4 m/s^2 

(i)  1400 = 0 +(1/2)*4*t^2   so  t = sqrt(700) secs  ≈ 26.5 secs

(ii) v = 0 + 4*t   so v = 4*sqrt(700) m/s  ≈ 106 m/s

A. 

B. You should be able to use the graph above to answer part B.

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that*

U can say hat the V is on the x-axis and v is the speed and the starting velocity is zero.

Is V the final velocity, the average velocity or what?  Is the starting velocity zero?

Having generated and stored your picture, select the Image icon in the menu bar of the input:

Then go to the Upload tab, browse to your picture and upload.

Then back to the Image tab and select OK.

Have a look at http://web2.0calc.com/questions/use-part-1-of-the-fundamental-theorem-of-calculus-to-find-the-derv#r1 and see if you can use the same step by step process given there.

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Using the expression due to Leibnitz, we have:

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according to my mathematics book its says that the answer is 10 seconds.

First I chose two of the equations: ab=a/b

(ab)/b = (a/b)/b

a = a/b^2

(a)/a = (a/b^2)/a

1 = a/b^2/a

1 = 1/b^2

b^2 = 1

b = sqrt(1)

b = 1

Plugged it back into the equations

a(1) = a + 1 = a/1

a = a + 1 = a/1

a = a + 1 is false

There is no solution.

The closest that's possible is 0, where:

(0)(0) = 0 + 0 = 0/0

0 = 0 = undefined

0/0 is undefined and thus there is no solution.

If you want reasoning for 0/0 being undefined:

It is understood that any number over itself equals 1 (such as 1/1 = 1, 2/2 = 1, 27/27 = 1)

It is also understood that zero divided by any number equals 0 (such as 0/1 = 1, 0/2 = 0, and 0/27 = 0)

Because both arguments stand, you cannot define 0/0 as either 1 or 0, leaving it as undefined.

Thank You for your help

6.86 approximately 7

First State Bridge-Painting Costs

\(\begin{array}{|rcll|} \hline \text{Painting Costs } = 1000\cdot \left[ -24 + \frac{351}{6} \cdot \left(\frac{\text{Length}}{100}\right) - 19 \cdot \left(\frac{\text{Length}}{100}\right)^2 + \frac{15}{6} \cdot \left(\frac{\text{Length}}{100}\right)^3 \right] \\ \hline \end{array}\)

3a
means

a + a + a

so 3a + a is the same as

a + a + a + a

4a

means

a + a + a + a

So the answer is 4a

Here is the graph of 

\(y=x^3-3x+4\)

You MUST understand this.

Now when you are integrating this function between 2 set values you are finding the area between the x axis the curve and those 2 vertical lines.

So look at this graph of    

\(y=x^3-3x+4\)

The integral is the green area under the curve.  

The minumum value is 2 and the maximum value is 22

You should be able to see that the area MUST be bigger than 2*(3-0) and smaller than   22*(3-0)

so

\(6<\int_0^3\;\;(x^3-3x+4)\;dx\;\;<66\)

\(6\;<\;\int_0^3\;(x^3-3x+4)\;dx\;<\;66\)

Understand that the integral sign S is exactly that - a stylized S which stands for 'sum of'  

You are summing the area of infintesimally narrow rectagles to find the total area under the curve.

mmm

If M and m are both positive then the integral will be between  (4-2)*m (smallest)  and (4-2)*M (largest)

If M or n can be negative and you do not know which one has the biggest absolute value then it gets more complicated.

Remember that an integral is the sum of infinitely tiny cross-section areas.