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2 years ago

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Which is the best way to solve the equation x(squared)=m? A) take the square root of both sides of the equation. B) multiply bot

h sides of the equation by m. C) square both sides of the equation. D) divide both sides of the equation by 2.

1 answer:

shepuryov [24]2 years ago

5 0

A
take square root of both sides

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Evaluate.<br> x+1/3<br> when x = -2.5<br> O 0.5<br> O -0.5<br> O 1.167<br> O-1

mario62 [17]

Answer:

-.5

Step-by-step explanation:

(x+1)/3

Let x = -2.5

(-2.5+1)/3

Parentheses first

-1.5/3

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2 years ago

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How does the graph of g(x) = −(x + 3)^4 compare to the parent function of f(x) = x^4

kotykmax [81]

Answer: Option D

g(x) is shifted 3 units to the left and reflected over the x-axis.

Step-by-step explanation:

If we have a main function $f (x) = x ^ 4$

And we perform the transformation:

$g (x) = f (x + h) = (x + h) ^ 4$

Then it is fulfilled that:

If $h> 0$ the graph of f(x) moves horizontally h units to the left

If $h$ the graph of f(x) moves horizontally h units to the right

If we have a main function $f (x) = x ^ 4$

And we perform the transformation:

$g (x) = -f(x) = -x ^ 4$

Then it is fulfilled that:

The graph of g(x) is equal to the graph of f(x) reflected on the x axis

In this case we have to:

$g(x) = -(x + 3)^4$ and $f(x) = x^4$

Therefore $h=3>0$ and $g(x) = -f(x)$

This mean that: g(x) is shifted 3 units to the left and reflected over the x-axis.

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2 years ago

you have $20 to spend on taxi fare. the ride costs $5 plus $2.50 per mile let m represent the number of miles ridden. write an i

Cerrena [4.2K]

<span>Let m represent the miles ridden. write an inequality to determine how many miles you can ride for $20.

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2 years ago

Determine the value(s) of k such that the system of linear equations has the indicated number of solutions. (Enter your answers

Ratling [72]

Answer:

k=3

Step-by-step explanation:

The given system of equations are

$9x+ky=24$

$kx+y=8$

We need to find the value of k if the given system of equation have infinite number of solutions.

If two equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ have infinite number of solutions, then

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

Using this we get

$\dfrac{9}{k}=\dfrac{k}{1}=\dfrac{-24}{-8}$

$\dfrac{9}{k}=\dfrac{k}{1}=3$

$\dfrac{k}{1}=3$

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Therefore, the value of k is 3.

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3 years ago

Compute the Taylor expansion of order n=2 of the function sin(xy) at x=0 and y=0

artcher [175]

Answer:

f(x, y) = Sin(x*y)

We want the second order taylor expansion around x = 0, y = 0.

This will be:

$f(x,y) = f(0,0) + \frac{df(0,0)}{dx} x + \frac{df(0,0)}{dy} y + \frac{1}{2} \frac{d^2f(0,0)}{dx^2} x^2 +\frac{1}{2} \frac{d^2f(0,0)}{dy^2}y^2 + \frac{d^2f(0,0)}{dydx} x*y$

So let's find all the terms:

Remember that:

$\frac{dsin(ax)}{dx} = a*cos(ax)$

$\frac{dcos(ax)}{dx} = -a*cos(ax)$

f(0,0) = sin(0*0) = 1.

$\frac{df(0,0)}{dx}*x = y*cos(0*0)*x = x*y$

$\frac{df(0,0)}{dy} *y = x*cos(00)*y = x*y$

$\frac{1}{2} \frac{d^2f(0,0)}{dx^2}*x^2 = -\frac{1}{2} *y^2*sin(0*0)*x^2 = 0$

$\frac{1}{2} \frac{d^2f(0,0)}{dy^2}*y^2 = -\frac{1}{2} *x^2*sin(0*0)*y^2 = 0$

$\frac{d^2f(0,0)}{dxdy} x*y = (cos(0*0) -x*y*sin(0*0))*x*y = x*y$

Then we have that the taylor expansion of second order around x = 0 and y = 0 is:

sin(x,y) = x*y + x*y + x*y = 3*x*y

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2 years ago