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3 years ago

Lenny’s favorite radio station has this schedule:

1 answer:

4 0

Answer: choice C) 19/60

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Explanation:

Add up the minute values: 19+4+37 = 60

We have a total of 60 minutes. Of that total, 19 minutes is dedicated to news.

Divide the two values (19 and 60) to get the fraction 19/60. We can't reduce from here so we leave it like that.

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Use the quadratic formula to solve the equation x^2-7x-6=0

olganol [36]

Answer:

$\frac{7\pm\sqrt{73}} {2}$

Explanation:

We have been given with the quadratic equation $x^2-7x-6=0$

We have general formula to find the roots of a quadratic equation first we find the discriminant with formula

$D=b^{2}-4ac$

and after that to find the variable suppose x we have the formula

$x=\frac{-b\pm\sqrt{D}} {2a}$

And general quadratic equation is

$ax^2+bx+c=0$

On comparing the given quadratic equation with genral quadratic equation we will have values

a=1, b=-7 and c=-6

After substituting these values in the formula we will get

$D=(-7)^2-4(1)(-6)={73}$

After substituting in the formula to find x we will get

$\frac{-(-7)\pm\sqrt{73}} {2}= \frac{7\pm\sqrt{73}} {2}$

4 0

4 years ago

Read 2 more answers

A state vector X X for a four-state Markov chain is such that the system is four times as likely to be in state 2 as in 4, is no

Vesna [10]

All the components in the state vector need to sum to 1. You're given that component corresponding to state 1 is 0.2, and that the component for state 3 is 0.

That leaves states 2 and 4, for which you're told that the component for state 2 is four times as large. If $x_i$ is the component for state $i$ , then you have

$x_1+x_2+x_3+x_4=1\iff 0.2+4x_4+0+x_4=1\implies5x_4=0.8\implies x_4=0.16$

which means $x_2=4x_4=0.64$ . So the state vector is $\mathbf x=(0.2,0.64,0,0.16)$ .

3 0

4 years ago

(a) Show that a differentiable function f decreases most rapidly at x in the direction opposite the gradient vector, that is, in

Sophie [7]

Answer:

Step-by-step explanation:

$\text{Show that a differentiable function f decreases most rapidly at x in the }$

$\text{direction opposite the gradient vector, that is, in the direction of}$ $-\bigtriangledown f(x)$ $\text{. Let}\ \theta \ \text{be the angle between} \bigtriangledown f(x) \ \text{and unit vector u. Then } D_u f = \mathbf{|\bigtriangledown f| \ cos \ \theta }}$

$\text{Since the minimum value of} \ \ \mathbf{cos \ \theta} \ \ is \mathbf{-1} \ \text{occuring \ for \ 0} \le \ \theta \ < 2x, \\ \\ when \ \theta = \mathbf{\pi} , \text{the mnimum value of} \ D_uf \ is} -|\bigtriangledown f|, \text{occuring when the direction of u is } \\ \\ \ \mathbf{the \ opposite \ of} \ \text{the direction of } \ \bigtriangledown f (assuming \ \bigtriangledown f\ is \ not \ zero)$

b) $\text{From part A:}$

$If \ f(x,y) = x^4y -x^2y^2 \ \ decreases \ fastest \ at \ the \point \ (2,-5)\\ \\ F(x,y) = x^4y -x^2y^3 \\ \\ f_x = \dfrac{df}{dx}= \dfrac{d}{dx}(x^4y-x^2y^3) \\ \\ f_x = \dfrac{df}{dx}= y4x^3 -2y^3x \\ \\ For(2,-5) \\ \\ f_x = (-5)4(2)^3 -2(-5)^3(2) \\ \\ \mathbf{ f_x = 340}$

$However; f_y = \dfrac{df}{dy} = \dfrac{d}{dy}(x^4y - x^2y^3) \\ \\ f_y = x^4 -3x^2y^2 \\ \\ Now, for (2, -5)\\ \\f_y = (2)^4 -3(2)^2(-5)^2 \\ \\ f_y = -284$