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Brilliant_brown [7]
3 years ago
10

Find the difference in the volume and total area of a cylinder with both a radius and height of1 r-1, h-1

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
4 0

Although I am not clear as to whether the radius is r—1 or otherwise, but if i work with the values i see then:

Volume — T. S. A.

= πr²h — (2πr²+2πrh)

So after the simplification, I substitute the values given,

r= r—1 and h=h—1

And the answer is...

π(r—1)(rh—3r—3h+5)!

You might be interested in
Use Gaussian Elimination to find an equation of a polynomial that passes through points A(-5,-3), B(-2,3). C(3,3), D(6,19). Indi
Marrrta [24]

Answer:

The polynomial equation that passes through the points is 2-\frac{2}{3}x+\frac{1}{12}x^{2}+\frac{1}{12}x^{3}

Step-by-step explanation:

Suppose you have a function y = f(x) which goes through these points

A(-5,-3), B(-2,3). C(3,3), D(6,19)

there is a polynomial P(x) of degree 3 which goes through these point.

We use the fact that <em>four distinct points will determine a cubic function.</em>

P(x) is the degree 3 polynomial through the 4 points, a standard way to write it is

P(x) = a+bx+cx^2+dx^3

Next replace the given points one by one, which leads to a system of 4 equations and 4 variables (namely a,b,c,d)

-3=a+b\cdot-5+c\cdot -5^2+d\cdot -5^3\\3=a+b\cdot-2+c\cdot -2^2+d\cdot -2^3\\3=a+b\cdot 3+c\cdot 3^2+d\cdot 3^3\\19=a+b\cdot 6+c\cdot 6^2+d\cdot 6^3

We can rewrite this system as follows:

-3=a-5\cdot b+25\cdot c-125\cdot d\\3=a-2\cdot b+4\cdot c-8\cdot d\\3=a+3\cdot b+9\cdot c+27\cdot d\\19=a+6\cdot b+36\cdot c+216\cdot d

To use the Gaussian Elimination we need to express the system of linear equations in matrix form (<em>the matrix equation Ax=b</em>).

The coefficient matrix (A) for the above system is

\left[\begin{array}{cccc}1&-5&25&-125\\1&-2&4&-8\\1&3&9&27\\1&6&36&216\end{array}\right]

the variable matrix (x) is

\left[\begin{array}{c}a&b&c&d\end{array}\right]

and the constant matrix (b) is

\left[\begin{array}{c}-3&3&3&19\end{array}\right]

We also need the augmented matrix, it is obtained by appending the columns of the coefficient matrix and the constant matrix.

\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\1&-2&4&-8&3\\1&3&9&27&3\\1&6&36&216&19\end{array}\right]

To transform the augmented matrix to the reduced row echelon form we need to follow these steps:

  • Subtract row 1 from row 2 \left(R_2=R_2-R_1\right)

\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&3&-21&117&6\\1&3&9&27&3\\1&6&36&216&19\end{array}\right]

  • Subtract row 1 from row 3 \left(R_3=R_3-R_1\right)

\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&3&-21&117&6\\0&8&-16&152&6\\1&6&36&216&19\end{array}\right]

  • Subtract row 1 from row 4 \left(R_4=R_4-R_1\right)

\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&3&-21&117&6\\0&8&-16&152&6\\0&11&11&341&22\end{array}\right]

  • Divide row 2 by 3 \left(R_2=\frac{R_2}{3}\right)

\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&1&-7&39&2\\0&8&-16&152&6\\0&11&11&341&22\end{array}\right]

  • Add row 2 multiplied by 5 to row 1 \left(R_1=R_1+\left(5\right)R_2\right)

\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&8&-16&152&6\\0&11&11&341&22\end{array}\right]

  • Subtract row 2 multiplied by 8 from row 3 \left(R_3=R_3-\left(8\right)R_2\right)

\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&0&40&-160&-10\\0&11&11&341&22\end{array}\right]

  • Subtract row 2 multiplied by 11 from row 4 \left(R_4=R_4-\left(11\right)R_2\right)

\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&0&40&-160&-10\\0&0&88&-88&0\end{array}\right]

  • Divide row 3 by 40 \left(R_3=\frac{R_3}{40}\right)

\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&0&1&-4&-1/4\\0&0&88&-88&0\end{array}\right]

  • Add row 3 multiplied by 10 to row 1 \left(R_1=R_1+\left(10\right)R_3\right)

\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&-7&39&2\\0&0&1&-4&-1/4\\0&0&88&-88&0\end{array}\right]

  • Add row 3 multiplied by 7 to row 2 \left(R_2=R_2+\left(7\right)R_3\right)

\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&88&-88&0\end{array}\right]

  • Subtract row 3 multiplied by 88 from row 4 \left(R_4=R_4-\left(88\right)R_3\right)

\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&0&264&22\end{array}\right]

  • Divide row 4 by 264 \left(R_4=\frac{R_4}{264}\right)

\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&0&1&1/12\end{array}\right]

  • Subtract row 4 multiplied by 30 from row 1 \left(R_1=R_1-\left(30\right)R_4\right)

\left[\begin{array}{cccc|c}1&0&0&0&2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&0&1&1/12\end{array}\right]

  • Subtract row 4 multiplied by 11 from row 2 \left(R_2=R_2-\left(11\right)R_4\right)

\left[\begin{array}{cccc|c}1&0&0&0&2\\0&1&0&0&-2/3\\0&0&1&-4&-1/4\\0&0&0&1&1/12\end{array}\right]

  • Add row 4 multiplied by 4 to row 3 \left(R_3=R_3+\left(4\right)R_4\right)

\left[\begin{array}{cccc|c}1&0&0&0&2\\0&1&0&0&-2/3\\0&0&1&0&1/12\\0&0&0&1&1/12\end{array}\right]

From the reduced row-echelon form the solutions are:

\left[\begin{array}{c}a=2&b=-2/3&c=1/12&d=1/12\end{array}\right]

The polynomial P(x) is:

2-\frac{2}{3}x+\frac{1}{12}x^{2}+\frac{1}{12}x^{3}

We can check our solution plotting the polynomial and checking that it passes through the points.

3 0
3 years ago
P + 0.20p as an equivalent expression
andreev551 [17]

Answer:

An equivalent expression to p - 0.20p is 0.8p Hope this helps :)

5 0
3 years ago
Just type the numerical answer.<br> Find the measure of angle A.<br> x+55<br> 86°<br> x+51
emmainna [20.7K]

Answer: x=-6

Step-by-step explanation:

x+55+x+51+86=180

2x+106+86=180

2x+192=180

2x=-12

x=-6

4 0
1 year ago
The area of a square shaped rug is 81 square feet if the rug is 9 feet long what is the perimeter
Yakvenalex [24]
81 x 9= 729 hope this helps
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Alexxx [7]

Answer:

y=mx+b is the final answer!

Step-by-step explanation:

i used a calculator so i am 100% sure these are right.

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