Two angles are complementary. The sum of the measure of the first angle and one-fourth the second angle is 73.5 degrees. Find the measure of the angles. Measure of the smaller angle is ?? and the measure of the other angle is??
that is exactly how the question is worded. I double-checked.
Thanks anyway...I'll skip this one and try it later.
x+(1/4)y=73.5 x+y=90 now solve
If this involves multiplying a fraction, I am lost already. Fractions are my enemy.
Oh, yeah, that makes sense
LOL...to YOU maybe
Well, its just two sentences that use the same variable. The first sentence says that the two angles are complementary, meaning they add to 90 degrees. You use x and y to represent both angles and their sum adding to 90 degrees. The second sentence says take a fourth of whatever the the second angle is, then add it to the first angle, and you will get 73.5 degrees.
Yea, but there are many different numbers you could use to make x + y = 90. How do you know which ones to use?
73.5/4 would be 18.375 as you said earlier.
No, you can't solve it like that. Forget what I did.
okay
This is what is called a systems of equations problem. You have to solve both equations for y, then set the equations equal to each other, then solve for x: y = 90 - x y = 4(73.5 - x) 90 - x = 4(73.5 - x) 90 - x = 294 - 4x 3x = 204 x = 68 y = 22
I am trying to understand what you did...
But I see how 68 and 22 = 90
I need to study what you did a little longer. So I am assuming the smaller angle is 22 degrees and the larger is 68 degrees?
1. The original equations were: x + y = 90 (because the angles are complementary) x + y/4 = 73.5 (because the second angle is only 1/4 of y and adds with x to get 73.5) 2. Solve both equations for y: y = 90 - x y = 4(73.5 - x) 3. Set y = y ( when you do the this, the first equation is set equal to the second and also the variable y is eliminated): 90 - x = 4(73.5 - x) 4. From here you just solve for x:
Had to fix something...
I think I am getting it slowly but surely. Thanks very much.
I can post a classroom to explain it if you want...
If I had the time I certainly would. I only have 3 hours to get this turned in and I only have 6 questions answered...
I can also help you with the other questions
okay, well that sounds good. Lord knows I can use the help. LOL...where do I go?
What is your email address? I can send you a link to the classroom. Once I have your email you can delete it.
This is so that pranksters don't try to join the class as well.
Okay, you can delete it now.
Did you send the link? It's not there yet.
Sorry, it takes a minute to set up the classroom. I hope you have a microphone or headphones
I don't know if I do or not. I don't have headphones and have never used the mic
Hmmm, well, we won't be able to do this then. You don't have speakers either?
Yes, I have speakers
Okay, well, you'll have to use those then. Maybe you can also get your mic to work
I may have to do this later anyway...my son is hungry and wants me to fix him something to eat.
I will try to get back soon. I have to get this turned in tonight!!
I will post the link either way when I'm done setting it up
Check your email
ok, thanks
Okay, you'll have to post your email again, sorry. I had to completely re-set up my account and I didn't save your email sorry.
I have the link now.
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