Weekly Math Problem #82

Geometric Series. This week's problem is a second part to last week's problem, WMP #81. Since we looked at a geometric sequence last week, why don't we look the corresponding geometric series this week? Well...we are. ✏️ 👍🏿

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

       ✔️     How to find the sum of the first set of numbers from a geometric sequence  

            

WMP #82 says to...

Happy solving!

Check back on Saturday, April 30th for the solution, which will be posted below ⬇️.

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✏️📓 Solution Time! 📓✏️

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And we're back! Of course, to find the sum of the first eight terms of our (or any) geometric series, we can write it down and find the sum manually...but...what if, instead of eight terms, we had had to find eighteen terms?! I don't want to write that many terms. And guess, what? We don't have to. There's a formula for that!

Here is the formula:

In our situation, we have the following:

Well, it looks like the sum of the first eight terms of the geometric series whose first term is 2 with a ratio of 3 is 6560.

Now, just for fun, let's find the sum manually to verify that we've used the formula correctly. (This will also help us to appreciate the formula more.) Here are the first eight numbers generated from last week's geometric sequence with its corresponding sum:

...And it's a match! 

▪️ Did you find this problem easy? Hard? Somewhere between easy and hard?

▪️ Leave a comment to let me know what you thought about this week's problem. 

Thank you for solving with me this week. ✏️
WMP! #83 is up next. 🤓

Cheers!

The Younge Lady

 

Weekly Math Problem #81

Geometric Sequence. It has been a loooong while since I have done any work with geometric sequences...or series. So, now is as good a time as any to do it. 🤷🏿‍♀️ ✏️

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

       ✔️     Understanding what a geometric sequence is
       ✔️     How to find the nth term in a geometric sequence

            

WMP #81 says to...

Happy solving!

Check back on Saturday, April 16th 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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Alright folks... I don't know about you, but I needed a way to organize the information; I need to see what I'm working with. So, I used a chart:

The chart shows what information I have, what information is missing, and helps me to see how I can approach the problem. From the chart, I can see that there are four factors (or ratios) between the second and the sixth terms. I can also see that I have enough information to figure out what r is.

Now that I know the common ratio is 3, I can use that information to find the first term. Why do I need to find the first term? Well...in the formula needed to find the n-th term, the first term is required. So, let's proceed to see what the first term is. That is done by taking the second term and dividing it by the common ratio of 3.

 

Cool...the first term, a1, is 2. It looks like I have enough information to find the 8th term. But...before I do that, I want to verify that the information I have--the common ratio of 3 and the first term, a1, of 2 will generate the 6th term. 

Nice! It works. With confidence, the problem can be completed. The 8th term can be found.

There you have it. The 8th term of a geometric sequence whose 2nd and 6th terms are 6 and 486, respectively, is 4,347.

▪️ Did you use the same methods above to solve the problem? If so, please share.

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

Thank you for solving with me this week. ✏️
WMP! #82 is up next. 🤓

Cheers!

The Younge Lady

Weekly Math Problem #80

Equation of a Circle. We're definitely keeping our algebra skills sharp with this week's problem. 🧠 As the saying goes, "Repetition is the mother of all learning". So, you will definitely see skills being repeated in the problems as well the combination of them to solve the problems. The equation of a circle is part of a larger topic called conic sections. Let's  go!  🏃🏿‍♀️

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

       ✔️     Substitution
       ✔️     Completing the square technique
       ✔️     How to solve a quadratic equation

            

WMP #80 wants us to...

Happy solving!

Check back on Saturday, April 9th 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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Part a.

To determine the center and radius of the circle, the completing the square technique needs to be used on the given equation. To start that process, the constant needs to be isolated. Once the equation is in standard form, then the center and radius can be identified.

The center of the circle is (1, -1) and its radius is 1 unit. This was figured out without graphing it. We'll view the graph later in this post.

Part b.

To show that the point (1, -2) is a point of intersection for circle C and the given line, the substitution method was used. The equation of the line was substituted for y in the equation of the circle. The equation of the circle was then simplified into a factorable quadratic equation. This allows for values of x to be solved for...one of which is the x-value in the given point. It is then shown that the corresponding y-value is indeed the y-value in the given point.

Now that we've completed both parts of the problem, let's view the graph:

**This plot was created using Geogebra's Graphing Calculator. Click image to enlarge.

 

▪️ Did you use the same methods above to solve the problem? If so, please share.

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

Thank you for solving with me this week. ✏️
Let's move on to WMP! #81. 💪🏿

Cheers!

The Younge Lady

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 ⬇️.

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 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