All categories

2 years ago

How do I do this number 5 and 6

Mathematics

1 answer:

grin007 [14]2 years ago

4 0

Answer:

5. 40

Step-by-step explanation:

(14) +(4x+6) = 8x

14 + 4x+ 6 = 8x

20 = 8x - 4x

4x = 20

x = 5

RT = 8x

= 8 * 5....since x = 5

= 40

You might be interested in

Find the slope of the line that goes through the points (-1,-2) and (-3,-4)

diamong [38]

Answer:

the answer is 1

(-4-(-2)) divided by (-3-(-1))

6 0

3 years ago

The following graphs show the performance of two stocks over the first five months of the year. A graph titled Stock A B C has m

hichkok12 [17]

Answer:

C) Stock XYZ shows the best performance. The scale of Stock XYZ makes the graph appear to be rising more slowly.

Step-by-step explanation:

i did this question on my test, from reading the graph, i am almost certain this is the answer. correct me if i am wrong!

8 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

The Cooking Club made some pies to sell

Alenkinab [10]

The club made 5 pies. If each pie is cut into 5 slices, and there are now 40 slices after the cafeteria donated, divide 40 by 5, which is 8. Subtract the 3 that the cafeteria donated.

3 0

3 years ago

What is the midpoint of the segment shown below?

olasank [31]

Answer: There all the same answers :|

Step-by-step explanation:

7 0

2 years ago