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DochEvi [55]
2 years ago
10

What does the following word mean? Floccinaucinihilipilification

Mathematics
1 answer:
vivado [14]2 years ago
3 0
<h3>Floccinaucinihilipilification means the action or habit of estimating something as worthless.</h3>
You might be interested in
PLEASE HELPPPPPJohn and Kathy are shopping for their first home. Between the two of them, they feel they can afford a $1400 mont
xxTIMURxx [149]

Answer: $249,300

Step-by-step explanation:

The exact formula would be if I think this is right

PV = PMT(1-(1+r)^-n)/r with r & n adjusted for periodicity

= 1400(1-(1+.054/12)^-360)/(.054/12)

= $249,318.47

which you can round off to, say, $249,300 <------

note:

counterchechecked with a financial calculator

5 0
2 years ago
How to do the inverse of a 3x3 matrix gaussian elimination.
nata0808 [166]

As an example, let's invert the matrix

\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}

We construct the augmented matrix,

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 2 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]

On this augmented matrix, we perform row operations in such a way as to transform the matrix on the left side into the identity matrix, and the matrix on the right will be the inverse that we want to find.

Now we can carry out Gaussian elimination.

• Eliminate the column 1 entry in row 2.

Combine 2 times row 1 with 3 times row 2 :

2 (-3, 2, 1, 1, 0, 0) + 3 (2, 1, 1, 0, 1, 0)

= (-6, 4, 2, 2, 0, 0) + (6, 3, 3, 0, 3, 0)

= (0, 7, 5, 2, 3, 0)

which changes the augmented matrix to

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]

• Eliminate the column 1 entry in row 3.

Using the new aug. matrix, combine row 1 and 3 times row 3 :

(-3, 2, 1, 1, 0, 0) + 3 (1, 1, 1, 0, 0, 1)

= (-3, 2, 1, 1, 0, 0) + (3, 3, 3, 0, 0, 3)

= (0, 5, 4, 1, 0, 3)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 5 & 4 & 1 & 0 & 3 \end{array} \right]

• Eliminate the column 2 entry in row 3.

Combine -5 times row 2 and 7 times row 3 :

-5 (0, 7, 5, 2, 3, 0) + 7 (0, 5, 4, 1, 0, 3)

= (0, -35, -25, -10, -15, 0) + (0, 35, 28, 7, 0, 21)

= (0, 0, 3, -3, -15, 21)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 3 & -3 & -15 & 21 \end{array} \right]

• Multiply row 3 by 1/3 :

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Eliminate the column 3 entry in row 2.

Combine row 2 and -5 times row 3 :

(0, 7, 5, 2, 3, 0) - 5 (0, 0, 1, -1, -5, 7)

= (0, 7, 5, 2, 3, 0) + (0, 0, -5, 5, 25, -35)

= (0, 7, 0, 7, 28, -35)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 0 & 7 & 28 & -35 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Multiply row 2 by 1/7 :

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Eliminate the column 2 and 3 entries in row 1.

Combine row 1, -2 times row 2, and -1 times row 3 :

(-3, 2, 1, 1, 0, 0) - 2 (0, 1, 0, 1, 4, -5) - (0, 0, 1, -1, -5, 7)

= (-3, 2, 1, 1, 0, 0) + (0, -2, 0, -2, -8, 10) + (0, 0, -1, 1, 5, -7)

= (-3, 0, 0, 0, -3, 3)

\left[ \begin{array}{ccc|ccc} -3 & 0 & 0 & 0 & -3 & 3 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Multiply row 1 by -1/3 :

\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

So, the inverse of our matrix is

\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}^{-1} = \begin{bmatrix}0&1&-1\\1&4&-5\\-1&-5&7\end{bmatrix}

6 0
2 years ago
Which of the following is true?
EleoNora [17]
The options can be rewrited as follow:
A. 5<4
B. 5<5
C. 5<4
D. 4<-5

∴ Answer is none of the above.

Unless perhaps you have written down the question incorrectly.
8 0
3 years ago
Is -2.22 a rational or real or integers
mezya [45]
All because -2.22 is a real number, it’s a rational number because it can stop and a integer because it’s negative
3 0
3 years ago
Read 2 more answers
¿Cuál es el volumen en pulgadas cúbicas de una esfera cuya circunferencia mide 64 pulgadas?
pochemuha

Answer:

4445.18 pulgadas cúbicas

Step-by-step explanation:

Paso 1

La fórmula para la circunferencia de una esfera = 2πr

Circunferencia de una esfera = 64 pulgadas

Por eso,

64 = 2πr

Por lo tanto, encontramos r

Dividir ambos lados por 2π

64 / 2π = 2πr / 2π

r = 10.185916358 pulgadas

Radio r = 10.2 pulgadas

Paso 2

Volumen de una esfera

= 4/3 × π × r³

r = 10.2 pulgadas

Por lo tanto, 4/3 × π × (10.2) ³

= 4445.18 pulgadas cúbicas

Por lo tanto, el volumen en pulgadas cúbicas de una esfera = 4445.18 pulgadas cúbicas

6 0
3 years ago
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