Weekly Math Problem #79

Trapezoid Angles. This is the first geometry problem of the year. Hopefully, it won't be the only one. I'm just switching things up a bit. I looked at my list of past problems and noticed that I didn't see the word geometry on it. Then, I just looked on the internet for geometry problems and landed on the one below.

To solve this week's problem in completion, you need to recall the following math skills and information:

       ✔️     Properties of a trapezoid
       ✔️     Properties of a triangle
       ✔️     Properties of parallel lines cut by a transversal

            

WMP #79 wants us to...

Happy solving!

Check back on Saturday, April 2nd for the solution, which will be posted below ⬇️.

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Buy Me a ☕️ Coffee: TheYoungeLady ( I'm gonna need it this year. 😆 )

✏️📓 Solution Time! 📓✏️

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Let's jump right in! The shape we're dealing with is a trapezoid. In a trapezoid, the bases are parallel. Even if you didn't know that property of a trapezoid, it is indicated in the image. Notice in the picture there are two bold triangles facing the same direction on the bases; this shows parallelism. This is so useful! Why? Well, the two parallel lines along with the diagonal, allow us to identify alternate interior angles. 😁 Alternate interior angles are congruent! In our case, the alternate interior angles we have allow us to solve for x.

Now that we've found the value of x, we can find the value of y. The x-value can be substituted into an equation while simultaneously solving for the y-value. How? What is it that allows us to do that? The two angles in the trapezoid that contain the variable y, are part of a triangle that also contains the angles that allowed us to solve for x. We can use the property of the sum of the angles of a triangle to solve for y. Recall that the sum of the angles of a triangle is 180°.

...And just like that, we've found the value of the two variables: x = 10, and y = 25. 

 

▪️ Did you remember the geometry properties mentioned above?

▪️ Let me know what you thought about this week's problem in the comments section. 

Thank you for solving with me this week. ✏️
Next up WMP! #80. 👍🏿

Cheers!

The Younge Lady

Weekly Math Problem #78

2nd Derivative. If you're anything like me, then you may treat finding the derivative of a function like solving a puzzle. I find a lot of puzzles fun...until I get annoyed at how long it takes me to do it, if it's taking me too long. 🙄 Typically, finding the first derivative of function is not too bad. Finding the higher order derivatives is where some of the challenge may come in. This depends on what type of function the previous derivative yields. This week, we'll just go up to the second derivative. (If you want to go higher, let me know.)

To solve this week's problem in completion, you need to recall the following math skills and information:

       ✔️     Differentiation rules
       ✔️     How to find a derivative

            

WMP #78 says ...

Happy solving!

Check back on Saturday, April 2nd for the solution, which will be posted below ⬇️.

Shameless 🔌 Plug: Follow me on Instagram @TheYoungeLady
Buy Me a ☕️ Coffee: TheYoungeLady ( I'm gonna need it this year. 😆 )

✏️📓 Solution Time! 📓✏️

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Here we go. If you remember how derivatives work, then you'll remember that you have to find the first derivative before finding the second derivative. So let's do that. To find the first derivative of the function, the product rule is required.

Applying the product rule to h(x) is as follows:

Now that we've found the first derivative, we can find the second derivative...which is that the problem asked us to find. The first derivative consists of two terms--each needing the product rule to find its individual derivative. I use a green plus sign "+" to show the two terms, and their respective product rules are below their respective  diagonal arrows.

After applying the product rule twice on the first derivative and combining like terms, we have the second derivative, h''(x). 

 

▪️ Did you get the same answer for your second derivative?

▪️ Let me know what you thought about this week's problem in the comments section. 

Thank you for solving with me this week. ✏️
We're on to WMP! #79. 🤓

Cheers!

The Younge Lady

Weekly Math Problem #77

Asymptotes. Oh man...I was "this" 🤏🏿 close to skipping this week's problem. Nonetheless, I pressed on. I was scrolling through a College Algebra Study Guide (from my personal archives) when I came across the question you will see below. When I read it, it thought to myself, "Do I remember how to find asymptotes?". 🤔 Then I recalled that there are some rules to remember when it comes to finding any existing asymptotes for rational functions. So, this topic is the focus of this week's WMP.

Check out WMP #11 where I did some previous work with asymptotes.

To solve this week's problem in completion, you need to recall the following math skills and information:

        ✔️     Rules for finding asymptotes 
        ✔️     How to solve linear equations

            

WMP #77 says ...

Happy solving!

Check back on Saturday, April 2nd for the solution, which will be posted below ⬇️.

Shameless 🔌 Plug: Follow me on Instagram @TheYoungeLady
Buy Me a ☕️ Coffee: TheYoungeLady ( I'm gonna need it this year. 😆 )

✏️📓 Solution Time! 📓✏️

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Let's start with the HA (horizontal asymptote). There are two ways you can find the HA. One way is to pay attention to the degrees of the numerator and denominator, then follow the rule. In our case, the degree of the numerator and denominator are the same (each is degree one ---> linear). When the degrees are the same, then the value for the equation of the line is the ratio of the leading coefficients.

Alternatively, you can evaluate the function at a very large value...say 100 or more. (The larger, the better, as long as your machine can handle it.) The result will yield the value or an an approximate value for the horizontal asymptote. Remember, the equation for a horizontal line is y = , f(x) = , etc.

As for the VA (vertical asymptote), all you need to do is set the denominator equal to zero and solve for x. Your solution is the equation of the VA.

No oblique asymptotes exist because our function doesn't fit the criteria for having any.

Here is a graph of the function:

 

** This plot was generated using Geogebra.org's calculator. **

▪️ Were you able to find the asymptotes?

▪️ Let me know what you thought about this week's problem in the comments section. 

Thank you for solving with me this week. ✏️
We're on to WMP! #78. 🤓

Cheers!

The Younge Lady