please solve itwho answer first and with all explanation I will mark as brainliest answer​

Answer:

1/2 is ur answer .

Explanation:

Answer:

In _______, notochord is present only in larval tail while in ________, it extends from head to tail region and is persistent throughout their life.

A

Cephalochordata and Urochordata

B

Urochordata and Cephalochordata

C

Vertebrata and Cephalochordata

D

Cephalochordata and Vertebrata

Open in App

OpenApp

Solution

The correct option is B Urochordata and Cephalochordata

Phylum Chordata is divided into three subphyla - Urochordata,

Cephalochordata and Vertebrata.

Urochordata is characterised by the presence of notochord only in larval tail.

In Cephalochordata, notochord extends from head to tail region and is persistent throughout their life.

The notochord of Vertebrata is replaced by a cartilaginous or bony vertebral column in the adult.

flag

Suggest Corrections

thumbs-up

1

Explanation:

follo me brinlist kr do ok

In the sub-phylum (_i_) notochord is present only in the larval tail, while in (_ii_), it extends from head to tail region and is persistent throughout their life.

Answer:

[tex]\qquad\qquad\boxed{ \sf{ \: \bf \: \: \left \{ \: \dfrac{ det(adj(adjA))}{7} \right \} = \dfrac{1}{7} \: }} \\ \\ [/tex]

Step-by-step explanation:

Given that,

[tex]\sf \:A = [ a_{ij}]_{4 \times 4} \: such \: that \\ \\ [/tex]

[tex]\begin{gathered}\begin{gathered}\bf\: a_{ij} = \begin{cases} &\sf{2, \: \: when \: i = j} \\ &\sf{0, \: \: when \: i \: \ne \: j} \end{cases}\end{gathered}\end{gathered} \\ \\ [/tex]

So, it means

[tex]\sf\implies \sf \: A=\left[\begin{array}{cccc}2&0&0&0\\0&2&0&0\\0&0&2&0\\0&0&0&2\end{array}\right] \\ \\ [/tex]

So,

[tex]\bf\implies \: |A| = 2 \times 2 \times 2 \times 2 = 16 \\ \\ [/tex]

Now, Consider

[tex]\sf \: |adj(adjA)| \\ \\ [/tex]

[tex]\sf \: = \: { |A| }^{(4 - 1)^{2} } \\ \\ [/tex]

[tex]\boxed{ \sf{ \: \because \: |adj(adjA)| = { |A| }^{ {(n - 1)}^{2} } \: }} \\ \\ [/tex]

[tex]\sf \: = \: { |A| }^{ {(3)}^{2} } \\ \\ [/tex]

[tex]\sf \: = \: {16}^{9} \\ \\ [/tex]

[tex]\sf \: = \: {( {2}^{4} )}^{9} \\ \\ [/tex]

[tex]\sf \: = \: {2}^{36} \\ \\ [/tex]

Now, Consider

[tex]\sf \: \: \left \{ \: \dfrac{ det(adj(adjA))}{7} \right \} \\ \\ [/tex]

[tex]\sf \: \: = \: \left \{ \: \dfrac{ |adj(adjA)| }{7} \right \} \\ \\ [/tex]

[tex]\sf \: \: = \: \left \{ \: \dfrac{ {2}^{36} }{7} \right \} \\ \\ [/tex]

[tex]\sf \: \: = \: \left \{ \: \dfrac{ {2}^{3 \times 12} }{7} \right \} \\ \\ [/tex]

[tex]\sf \: \: = \: \left \{ \: \dfrac{ {( {2}^{3}) }^{12} }{7} \right \} \\ \\ [/tex]

[tex]\sf \: \: = \: \left \{ \: \dfrac{ {( 8) }^{12} }{7} \right \} \\ \\ [/tex]

[tex]\sf \: \: = \: \left \{ \: \dfrac{ {(7 + 1) }^{12} }{7} \right \} \\ \\ [/tex]

[tex]\sf \: \: = \: \left \{ \: \dfrac{C_0 + C_1(7) + C_2( {7}^{2} ) + ... + C_{12}( {7}^{12})}{7} \right \} \\ \\ [/tex]

[tex]\sf \: \: = \: \left \{ \: \dfrac{1 +7[ C_1+ C_2( {7}^{} ) + ... + C_{12}( {7}^{11})]}{7} \right \} \\ \\ [/tex]

[tex]\sf \: \: = \: \left \{ \: \dfrac{1 }{7} +C_1+ C_2( {7}^{} ) + ... + C_{12}( {7}^{11})\right \} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{1}{7} \\ \\ [/tex]

Hence,

[tex]\sf\implies \bf \: \: \left \{ \: \dfrac{ det(adj(adjA))}{7} \right \} = \dfrac{1}{7} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

If A and B are square matrices of order n, then

[tex]\sf \: |AB| = |A| \: |B| \\ \\ [/tex]

[tex]\sf \: |adjA| = { |A| }^{n - 1} \\ \\ [/tex]

[tex]\sf \: |A \: adjA| = { |A| }^{n} \\ \\ [/tex]

[tex]\sf \: | {A}^{ - 1} | = { |A| }^{ - 1} \\ \\ [/tex]

[tex]\sf \: |A| = |A'| \\ \\ [/tex]

[tex]\sf \: |adj(adj \: A)| = { |A| }^{ {(n - 1)}^{2} } \\ \\ [/tex]

Answer:

हाँ सही कहा आपने

Explanation:

Thank you