Five eighths added to a number is 4

Mathematics

2 answers:

AnnyKZ [126]3 years ago

6 0

Answer:

Step-by-step explanation:

4 i think

True [87]3 years ago

3 0

Answer:

what?

Step-by-step explanation:

You might be interested in

Find the solutions of the system: y = 2(5)-x and x + y = 2

bixtya [17]

The two equations in y-intercept form are:
y=-x+2 and y=-x+10

You have to graph them to find the solution, which is the point they intercept each other.

Both lines have the same slope so they will never intercept which means they have no solution.

So, the answer is no solution.

I hope that helps!

7 0

3 years ago

I need help with these

muminat

Answer:

Step-by-step explanation:

are we solving this ?

3 0

4 years ago

(problem 83)

AVprozaik [17]

To find the derivative of this function, there is a property that we should know called the Constant Multiple Rule, which says:

$\dfrac{d}{dx}[cf(x)] = cf'(x)$ (where $c$ is a constant)

Remember that the derivative of $\csc(x)$ is $-\csc(x)\cot(x)$ . However, you may notice that we are finding the derivative of $\dfrac{1}{2}\csc(2x)$ , not $\dfrac{1}{2} \csc(x)$ . So, we are going to have to use the chain rule. To complete the chain rule for the derivative of a trigonometric function (in layman's terms) is basically the following: First, complete the derivative of the trig function as you would if what was inside the trig function is $x$ . Then, take the derivative of what's inside of the trig function and multiply it by what you found in the first step.

Let's apply that to our problem. Right now, I am not going to worry about the $\dfrac{1}{2}$ at the front of the equation, since we can just multiply it back in at the end of our problem. So, let's examine $\csc(2x)$ . We see that what's inside the trig function is $2x$ , which has a derivative of 2. Thus, let's first find the derivative of $\csc(2x)$ as if $2x$ was just $x$ and then multiply it by 2.

The derivative of $\csc(2x)$ would first be $-\cot(2x)\csc(2x)$ . Multiplying it by 2, we get our derivative of $-2\cot(2x)\csc(2x)$ . However, don't forget to multiply it by the $\dfrac{1}{2}$ that we removed near the beginning. This gives us our final derivative of $-\cot(2x)\csc(2x)$ .

Remember that we now have to find the derivative at the given point. To do this, simply "plug in" the point into the derivative using the x-coordinate. This is shown below:

$-\cot[2(\dfrac{\pi}{4})]\csc[2(\dfrac{\pi}{4})]$