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4 years ago

What's the algebraic expression for 5,9,13,17,21

1 answer:

5 0

1,5,9,13,17,21,25,29..., 4n+1
in general terms of an arithmetic sequence with the first term A0 and common differences d,

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If the denominator of 5/7 is increased by a number and the numerator is doubled, the result is 1. Find the number.

iren2701 [21]

Answer:

Step-by-step explanation:

let n be the number, then

$\frac{10}{7+n}$ = 1 ← multiply both sides by 7 + n

10 = 7 + n ( subtract 7 from both sides )

3 = n

6 0

3 years ago

Help me with this please please pleasee

Mazyrski [523]

Answer:

37.79924

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Select all statements below which are true for all invertible n×n matrices A and B

Mekhanik [1.2K]

Answer:

a. False

b. False

c. True

d. True

e. False

f. False

Step-by-step explanation:

Hi,

We have certain properties for matrices,<em> (where A and B are nxn matrices and I is the identity matrix) </em>:

${(A^{-1})}^{-1} = A$

$(AB)^{-1} = B^{-1}A^{-1}$

$(A')^{-1} = (A^{-1})'$

$(A^{n})^{-1} = (A^{-1})^{n} = A^{-n}$

$AA^{-1} = A^{-1}A = I$

$AI = IA = A$

Using these properties, we verify the provided statements:

A. False.

None of the properties help verify this statement. We ca use an example for counter:

Let $A =\left[\begin{array}{cc}1&2\\2&0\\\end{array}\right]$ and $B = \left[\begin{array}{cc}5&1\\3&2\end{array}\right]$ , we calculate the L.H.S:

$A+B = \left[\begin{array}{cc}1+5&2+1\\2+3&0+2\end{array}\right]\\= \left[\begin{array}{cc}6&3\\5&2\end{array}\right]$

The square of (A+B):

$(A + B)^{2} = \left[\begin{array}{cc}36&9\\25&4\end{array}\right]$

Lets calculate the R.H.S:

$A^{2} =\left[\begin{array}{cc}1&4\\4&0\\\end{array}\right]\\B^{2} = \left[\begin{array}{cc}25&1\\9&4\end{array}\right]\\2AB = \left[\begin{array}{cc} (1 \times 5) + (2 \times 3) &(1 \times 1) + (2 \times 2)\\(2 \times 5) + (0 \times 3)& (2 \times 1) + (0 \times 2)\end{array}\right]\\= \left[\begin{array}{cc} 11 &5\\10& 2\end{array}\right]$

$A^{2} + B^{2} + 2AB = \left[\begin{array}{ccc}1+25+11&4+1+5 \\4+9+10&0+4+2\\\end{array}\right] \\= \left[\begin{array}{ccc}37&10 \\23&6\\\end{array}\right]$

This proves that: L.H.S ≠ R.H.S

Hence, A is false.

B. False

This can only hold when the eigenvalues for A are real.

$trace (A^{2}) > 0, det (A^{2}) > 0 : \\(A + A^{-1}) = ( I + A^{2} ) A^ {- 1} = ( A ( I + A ^{2} )^ {-1})^ {-1}$

C. True

This is a simplification of the distribution property of matrices.

D. True

The property that inverse is possible for any "n" value of the matrix.

E. False

Similar to part A, we can show that this property is invalid for any nxn matrix. Let:

$A = \left[\begin{array}{cc}1&2\\0&1\end{array}\right] \\A^{-1} = \left[\begin{array}{cc}1&-2\\0&1\end{array}\right]$

L.H.S:

$A + A^{-1} = \left[\begin{array}{cc}1+1&2-2\\0+0&1+1\end{array}\right] = \left[\begin{array}{cc}2&0\\0&2\end{array}\right]$

$(A + A^{-1})^{9} = \left[\begin{array}{cc}512&0\\0&512\end{array}\right]$

R.H.S:

$A^{9} = \left[\begin{array}{cc}1&512\\0&1\end{array}\right] \\A^{-9}= \left[\begin{array}{cc}1&-0.001953125\\0&1\end{array}\right]\\\\\\A^{9} + A^{-9} = \left[\begin{array}{cc}2&512\\0&2\end{array}\right] \\$

Since, L.H.S ≠ R.H.S, the statement is false.

F. False

This is a basic matrix rule, that commutative property does not apply on matrices.

3 0

3 years ago

Line segment YV of rectangle YVWX measures 24 units.

oksian1 [2.3K]

Answer: $YX=13.85\ units$

Step-by-step explanation:

<h3> The complete exercise is: "Line segment YV of rectangle YVWX measures 24 units. What is the length of line segment YX?"</h3><h3> The missing figure is attached.</h3>

Since the figure is a rectangle, you know that:

$WV=YX\\\\YV=XW=24$

Notice that the segment YV divides the rectangle into two equal Right triangles.

Knowing the above, you can use the following Trigonometric Identity:

$tan\alpha =\frac{opposite}{adjacent}$

You can identify that:

$\alpha =30\°\\\\opposite=YX\\\\adjacent=YV=24$

Therefore, in order to find the length of the segment YX, you must substitute values into $tan\alpha =\frac{opposite}{adjacent}$ and then you must solve for YX.

You get that this is:

$tan(30\°)=\frac{YX}{24}\\\\(24)(tan(30\°))=YX\\\\YX=13.85$

7 0

3 years ago

Read 2 more answers

Simplify (-2x)4<br><br> A. -16x4<br> B. -8x4<br> C. -2x4<br> D. 8x4<br> E. 16x4

Elan Coil [88]

Answer will E I believe

3 0

3 years ago