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ratelena [41]
3 years ago
12

Select all names that apply to the Number √4

Mathematics
2 answers:
Alexxandr [17]3 years ago
8 0
<h2><u>ANSWER</u><u>✔</u></h2>

\sf\therefore \sqrt{4}=?

\sf\implies \sqrt{ 2 \times2}

\sf\implies \sqrt{2^2}

\sf\implies 2

\large{\boxed{\bf{ \implies \:2\:}}}

<h2><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u></h2>
never [62]3 years ago
7 0

Answer:

\sqrt{4}  = 2

Step by step explaination:

\sqrt{4}   \\  =  \sqrt{ {2}^{2} }  \\  = 2

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Check the true statements below:
valentinak56 [21]

Answer:

a) False

b) False

c) True

d) False

e) False

Step-by-step explanation:

a. A single vector by itself is linearly dependent. False

If v = 0 then the only scalar c such that cv = 0 is c = 0. Hence, 1vl is linearly independent. A set consisting of a single vector v is linearly dependent if and only if v = 0. Therefore, only a single zero vector is linearly dependent, while any set consisting of a single nonzero vector is linearly independent.

b. If H= Span{b1,....bp}, then {b1,...bp} is a basis for H. False

A sets forms a basis for vector space, only if it is linearly independent and spans the space. The fact that it is a spanning set alone is not sufficient enough to form a basis.

c. The columns of an invertible n × n matrix form a basis for Rⁿ. True

If a matrix is invertible, then its columns are linearly independent and every row has a pivot element. The columns, can therefore, form a basis for Rⁿ.

d.  In some cases, the linear dependence relations among the columns of a matrix can be affected by certain elementary row operations on the matrix. False

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3 0
3 years ago
Write the function a a product of linear factor by grouping or using the x method or a combination of both
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<h3><u>Answer:</u></h3>

\boxed{\boxed{\pink{\sf Option \ A \ is \ correct .}}}

<h3><u>Step-by-step explanation:</u></h3>

Given function to us is :-

\bf \implies g(x) = x^2 - 9

And we , need to write the function a a product of linear factor by grouping or using the x method or a combination of both . So let's factorise this ,

\bf \implies g(x) = x^2 - 9 \\\\\bf\implies g(x) = x^2-3^2\\\\\bf\implies \boxed{\red{\bf g(x) = (x+3)(x-3) }}\:\:\bigg\lgroup \blue{\tt Using \ (a+b)(a-b) \ = a^2-b^2 }\bigg\rgroup

I have also attached the graph of x²-9.

<h3><u>Hence </u><u>option</u><u> </u><u>A</u><u> </u><u>is</u><u> </u><u>corr</u><u>ect</u><u> </u><u>.</u></h3>

4 0
3 years ago
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