What is a variable?

Mathematics

2 answers:

Vanyuwa [196]3 years ago

4 0

Answer:

B. A variable is something we do not know and we use a letter to represent it.

velikii [3]3 years ago

3 0

Answer:

C. A variable is the number in front of the letter in an expression.

You might be interested in

Write the given second order equation as its equivalent system of first order equations. u′′−5u′−4u=1.5sin(3t),u(1)=1,u′(1)=2.5

lakkis [162]

Answer:

hi your question options is not available but attached to the answer is a complete question with the question options that you seek answer to

Answer: v = 5v + 4u + 1.5sin(3t),

1.5sin(3t)

Step-by-step explanation:

u" - 5u' - 4u = 1.5sin(3t) where u'(1) = 2.5 u(1) = 1

v represents the "velocity function" i.e v = u'(t)

As v = u'(t)

<em>u' = v</em>

since <em>u' = v </em>

v' = u"

v' = 5u' + 4u + 1.5sin(3t) ( given that u" - 5u' - 4u = 1.5sin(3t) )

= 5v + 4u + 1.5sin(3t) ( noting that v = u' )

so v' = 5v + 4u + 1.5sin(3t)

d/dt $\left[\begin{array}{ccc}u&\\v&\\\end{array}\right]$ = $\left[\begin{array}{ccc}0&1&\\4&5&\\\end{array}\right]$ $\left[\begin{array}{ccc}u&\\v&\\\end{array}\right]$ + $\left[\begin{array}{ccc}0&\\1.5sin(3t)&\\\end{array}\right]$

Given that u(1) = 1 and u'(1) = 2.5

since v = u'

v(1) = 2.5

note: the initial value for the vector valued function is given as

$\left[\begin{array}{ccc}u(1)&\\v(1)\\\end{array}\right]$ = $\left[\begin{array}{ccc}1\\2.5\\\end{array}\right]$

7 0

3 years ago

Which simplifications of the powers of i are correct? There may be more than one correct answer.

fredd [130]

$\bf i^2=-1\qquad\qquad i^3=-i\qquad \qquad i^4=1 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ i^{22}\implies i^{(4\cdot 5)+2}\implies i^{4\cdot 5}i^2\implies (i^4)^5 i^2\implies 1^5(-1)\implies -1~\dotfill \bigotimes \\\\\\ i^{11}\implies i^{(2\cdot 5)+1}\implies (i^2)^5 i\implies (-1)^5(i)\implies -i~\dotfill \checkmark$

$\bf i^{21}\implies i^{(4\cdot 5)+1}\implies (i^4)^5 i\implies 1^5(i)\implies i~\dotfill \checkmark \\\\\\ i^{12}\implies i^{3\cdot 4}\implies i^3 i^4\implies (-i)(1)\implies -i\dotfill \bigotimes \\\\\\ i^{20}\implies i^{4\cdot 5}\implies (i^4)^5\implies 1~\dotfill \checkmark \\\\\\ i^{26}\implies i^{(4\cdot 6)+2}\implies (i^4)^6 i^2\implies 1^6(-1)\implies -1\dotfill \checkmark \\\\\\ i^{27}\implies i^{(4\cdot 6)+3}\implies (i^4)^6 i^3 \implies 1^6(-i)\implies -i\dotfill \bigotimes$