All categories

3 years ago

(9x−2.25)=(1.6+9x) help

1 answer:

4 0

Begin solving like you normally would. Add 2.25 to both sides and subtract 9x from both sides to try getting all x values on one side. However, you will find that after subtracting 9x from both sides, all x values go away, and you end with -2.25=1.6. This is not true (-2.25 is not 1.6), so the answer is no solution. There is no value of x which satisfies that equation.

You might be interested in

The sum of the two shorter sides of a triangle must be equal to the longest side to be classified as a triangle

kogti [31]

Answer:

False

Step-by-step explanation:

For example if you follow the pythagoras theorem a triangle has sides of 3,4 and 5

but when you add 3 and 4 they do not give you 5

so that is false

5 0

3 years ago

In Exercises 11-18, use analytic methods to find the extreme values of the function on the interval and where they occur. Identi

Colt1911 [192]

Answer:

Absolute maximum of 1 at x = pi/4 ; $(\frac{\pi}{4}, \ 1)$

Absolute minimum of -1 at x = 5pi/4 ; $(\frac{5\pi}{4} , \ -1)$

Local maximum of √2/2 at x = 0 ; $(0, \ \frac{\sqrt{2} }{2} )$

Local minimum of 0 at x = 7pi/4 ; $(\frac{7\pi}{4}, \ 0)$

No critical points that are not stationary points.

Step-by-step explanation:

$f(x)=sin(x+\frac{\pi}{4} ), \ 0 \leq x\leq \frac{7 \pi}{4}$

<h2>Take Derivative of f(x):</h2>

Let's start by taking the derivative of the function.

Use the power rule and the chain rule to take the derivative of f(x).

$f'(x)=\frac{d}{dx} [sin(x+\frac{\pi}{4})] \times \frac{d}{dx} (x+\frac{\pi}{4})$

The derivative of sin(x) is cos(x), so we can write this as:

$f'(x)=cos(x+\frac{\pi}{4})\times \frac{d}{dx} (x+\frac{\pi}{4})$

Now, we can apply the power rule to x + pi/4.

$f'(x)=cos(x+\frac{\pi}{4} ) \times 1$

$f'(x)=cos(x+\frac{\pi}{4} )$

<h2>Critical Points: Set f'(x) = 0</h2>

Now that we have the first derivative of $f(x)=sin(x+\frac{\pi}{4})$ , let's set the first derivative to 0 to find the critical points of this function.

$0=cos(x+\frac{\pi}{4})$

Take the inverse cosine of both sides of the equation.

$cos^-^1(0) = cos^-^1[cos(x+\frac{\pi}{4})]$

Inverse cosine and cosine cancel out, leaving us with x + pi/4. The inverse cosine of 0 is equal to 90 degrees, which is the same as pi/2.

$\frac{\pi}{2} = x +\frac{\pi}{4}$

Solve for x to find the critical points of f(x). Subtract pi/4 from both sides of the equation, and move x to the left using the symmetric property of equality.

$x=\frac{\pi}{2}- \frac{\pi}{4}$

$x=\frac{2 \pi}{4}-\frac{\pi}{4}$
$x=\frac{\pi}{4}$

Since we are given the domain of the function, let's use the period of sin to find our other critical point: 5pi/4. This is equivalent to pi/4. Therefore, our critical points are:

$\frac{\pi}{4}, \frac{5 \pi}{4}$

<h2>Sign Chart(?):</h2>

Since this is a sine graph, we don't need to create a sign chart to check if the critical values are, in fact, extreme values since there are many absolute maximums and absolute minimums on the sine graph.

There will always be either an absolute maximum or an absolute minimum at the critical values where the first derivative is equal to 0, because this is where the sine graph curves and forms these.

Therefore, we can plug the critical values into the original function f(x) in order to find the value at which these extreme values occur. We also need to plug in the endpoints of the function, which are the domain restrictions.

Let's plug in the critical point values and endpoint values into the function f(x) to find where the extreme values occur on the graph of this function.

<h2>Critical Point Values:</h2>

$f(\frac{\pi}{4} )=sin(\frac{\pi}{4} + \frac{\pi}{4} ) \\ f(\frac{\pi}{4} )=sin(\frac{2\pi}{4}) \\ f(\frac{\pi}{4} )=sin(\frac{\pi}{2}) \\ f(\frac{\pi}{4} )=1$

There is a maximum value of 1 at x = pi/4.

$f(\frac{5\pi}{4} )=sin(\frac{5\pi}{4} + \frac{\pi}{4} ) \\ f(\frac{5\pi}{4} )=sin(\frac{6\pi}{4}) \\ f(\frac{5\pi}{4}) = sin(\frac{3\pi}{2}) \\ f(\frac{5\pi}{4} )=-1$

There is a minimum value of -1 at x = 5pi/4.

<h2>Endpoint Values:</h2>

$f(0) = sin((0) + \frac{\pi}{4}) \\ f(0) = sin(\frac{\pi}{4}) \\ f(0) = \frac{\sqrt{2} }{2}$

There is a maximum value of √2/2 at x = 0.

$f(\frac{7\pi}{4} ) =sin(\frac{7\pi}{4} +\frac{\pi}{4}) \\ f(\frac{7\pi}{4} ) =sin(\frac{8\pi}{4}) \\ f(\frac{7\pi}{4} ) =sin(2\pi) \\ f(\frac{7\pi}{4} ) =0$

There is a minimum value of 0 at x = 7pi/4.

We need to first compare the critical point values and then compare the endpoint values to determine whether they are maximum or minimums.

<h2>Stationary Points:</h2>

A critical point is called a stationary point if f'(x) = 0.

Since f'(x) is zero at both of the critical points, there are no critical points that are not stationary points.

6 0

3 years ago

Can somebody solve this?

lisabon 2012 [21]

Answer:

x= 2

Step-by-step explanation:

Isolate the radical

Raise each side of the equation to the power of 3

5 0

3 years ago

Read 2 more answers

Prove sin(x-y)/cosxsiny=tanxcoty-1

Leviafan [203]

Answer:

See below

Step-by-step explanation:

Here I attach you the detailed solution.

For this we will need to use the following trigonometric properties and formulas:

sin (x-y) = sin (x) cos (y) - cos (x) sin (y)

tan(x) = sin(x)/cos(x)

cot(x) = 1/tan(x) = [tan(x)] ^(-1) = cos (x)/sin(x)

This properties can be found in any mathematics book and is valid to apply them here, and are the only tools we use to solve the problem. Its worth to note that we have to assume that cos(x)sin(y) is different from 0 in order to use it as a denominator.

8 0

4 years ago

I would greatly appreciate if someone could help me with this

Sergeu [11.5K]

Answer:

Number with 1 dog, number with 3, and number with 4 are all incorrectly plotted.

6 0

3 years ago