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3 years ago

Solve 0= 2t^3-21t^2+ 40t

1 answer:

4 0

$0=2t^3-21^2+40t$
switch sides
$2t^3-21t^2+40t=0$
solve factoring
$t(t-8)(2t-5)=0$
using the zero factor principle
$t=0
$
solve
$t-8=0:t=8$
solve
$2t-5=0:t= \frac{5}{2}$
$t=0,t=8,t= \frac{5}{2}$
Hope this helps

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The following are the properties of the sin trigonometric ratio of the function;

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From the above listed deductions, we can see that the true statement about the function y = sin x is that the range which is always known as the maximum and minimum values of the function are 1 and - 1 respectively.

Thus, the true statement relating to a property of the function y =sin x is that the maximum and minimum values of the function are 1 and -1 respectively. Option B

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Find an explicit rule for the nth term of the sequence.The second and fifth terms of a geometric sequence are 18 and 144, respec

vova2212 [387]

Given:

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Hence, we have:

$\begin{gathered} \text{ar}^{2-1}\text{ = 18} \\ ar\text{ = 18} \\ \\ ar^{5-1}=\text{ 144} \\ ar^4\text{ =144} \end{gathered}$

Divide the expression for the fifth term by the expression for the second term:

$\begin{gathered} \frac{ar^4}{ar}\text{ = }\frac{144}{18} \\ r^3\text{ = }\frac{144}{18} \\ r\text{ = 2} \end{gathered}$

Substituting the value of r into any of the expression:

$\begin{gathered} ar\text{ = 18} \\ a\text{ }\times\text{ 2 = 18} \\ Divide\text{ both sides by 2} \\ \frac{2a}{2}\text{ =}\frac{18}{2} \\ a\text{ = 9} \end{gathered}$

Hence, the explicit rule for the sequence is:

$a_n\text{ = 9\lparen2\rparen}^{n-1}$

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anyanavicka [17]

Answer:

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Difference= 456,14 seconds /60=7 minutes 36 seconds

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5.50 min =350 seconds

Time² =Speed¹ x Time¹ /Speed²

Time² =Speed¹ x (Time² + 350 seconds ) /Speed²

Time² =0.9 x (Time² + 350 seconds ) /1.9

Time² =0,4737 x (Time² + 350 seconds )

Time² =0,4737 Time² + 165,79 seconds

Time² - 0,4737 Time² = 165,79 seconds

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