Answer:
<em><u>The</u></em><em><u> </u></em><em><u>answer</u></em><em><u> </u></em><em><u>is</u></em><em><u>,</u></em><em><u> </u></em><em><u>q</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>16</u></em><em><u>.</u></em>
Step-by-step explanation:
1) Divide both sides by 3.
2) Simplify 27/3 to 9.
3) Add 7 to both sides.
4) Simplify 9 + 7 to 16.
<em><u>Therefor</u></em><em><u>,</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>answer</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>q</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>16</u></em><em><u>.</u></em>
56400 ÷ 0.332
<span>169879 total people
</span>
Hope this helps!
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),
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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Du%26%5C%5Cv%26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}0&1&\\4&5&\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%26%5C%5C4%265%26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
+ ![\left[\begin{array}{ccc}0&\\1.5sin(3t)&\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26%5C%5C1.5sin%283t%29%26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Du%281%29%26%5C%5Cv%281%29%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}1\\2.5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C2.5%5C%5C%5Cend%7Barray%7D%5Cright%5D)
They can be expressed as fractions
they have positive values and can be expressed as terminating decimals.
Answer:
Not equivalent
Step-by-step explanation:
5:3
15:12 simplified = 5:4
