Choose the correct answer for each question.

Which of the two functions below has the smallest minimum y-value?

dalvyx [7]

Answer:

They have no minimum y -value

3 0

3 years ago

Ln a+3 + ln a-3 = ln 16

Lilit [14]

$\bf log_{{ a}}(xy)\implies log_{{ a}}(x)+log_{{ a}}(y)\\\\
and\qquad \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-----------------------------\\\\
ln(a+3)+ln(a-3)=ln(16)\implies ln[(a+3)(a-3)]=ln(16)
\\\\\\
ln[a^2-3^2]=ln(16)\impliedby \textit{removing ln() from both sides}
\\\\\\
a^2-9=16\implies a^2=16+9\implies a^2=25
\\\\\\
a=\pm\sqrt{25}\implies a=\pm 5$

7 0

3 years ago

What method(s) would you choose to solve the equation: x2 + 2x - 6 = 0

Nataly_w [17]

Complete the square since can't factor
2/2=1
add 1 to both sides
x²+2x+1-6=1
factor perfect square
(x+1)²-6=1
add 6 to both sides
(x+1)²=7
sqrt both sides
x+1=+/-√7
minus 1 both sides
x=-1+/-√7

so x=-1+√7 or x=-1-√7

7 0

3 years ago

Read 2 more answers

According to Professor Robinson, which of the following aspects of A Streetcar Named Desire can be said to reflect elements of T

natita [175]

The best and most correct answer among the choices provided by the first question is letter E which is all of the above.

On the other hand, the best and most correct answer among the choices provided by the second question is letter A which is rebirth and coming death.

I hope my answer has come to your help. Have a nice day ahead and may God bless you always!

7 0

3 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

vodka [1.7K]

Answer:

a) $v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

$x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10$

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals:

$0< t$

For this case if we select two values let's say 0.25 and 0.5 we see that

$v(0.25) =6.1875, v(0.5)=3.75$

And we see that for $a=0.5 >0.25=b$ we have that f(b)>f(a) so then the function is decreasing on this case.

$1$

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

$v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2$

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

$2$

We see that the function would be increasing.

$3$

For this interval we will see that for any two points a,b with $a>b$ we have $f(a)>f(b)$ for example let's say a=3 and b =4

$f(a=3) =0 , f(b=4) =9 , f(b)>f(a)$

The particle is moving to the right then the velocity is positive so then the answer for this case is: $0$

Part c

$a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

5 0

4 years ago