Trigonometry (Functions and Graphs)

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edd fedds

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I had a study guide of 35 questions and I got kinda confused on these questions. I had used the formulas and got answers but some didn't quite match.
Really need help.

Use the quadratic formula to solve y2+10y+24 = 0.
a.
b. 4, 6
c.
d. -4, -6
*Answer: I used the quadratic formula to solve and got y=-6,-4. So would my answer be D.?

Solve by factoring: a2 -5a-14 = 0.
a. 5, 14
b. -7, 2
c. 2, 7
d. 7, -2
*Answer: I factored and got a=-2,7 so would my answer be D.?

Solve x2-2x-3>0.
a. (-3, 1)
b. (-1, 3)
c. (- ? , -3) U (1, ? )
d. (- ? , -1) U (3, ? )
*My answer was x<-1 or x>3 so would my answer be B. or D.?

Solve the inequality .
a. [-3/2, 3]
b. (- ? , -3] U [3/2, ? )
c. (- ? , -3/2] U [3, ? )
d. [-3, 3/2]
* My answer was -3/2<x<3 so would I choose C.?

Determine the center and radius of the circle (x+2)2+(y-1)2 = 16
a. center: (2, -1); radius: 4
b. center: (2, -1); radius: 16
c. center: (-2, 1); radius: 4
d. center: (-2, 1); radius: 16
* My answer was C. Am I right?
 
edd fedds said:
I had a study guide of 35 questions and I got kinda confused on these questions. I had used the formulas and got answers but some didn't quite match.
Really need help.

Use the quadratic formula to solve y2+10y+24 = 0.
a.
b. 4, 6
c.
d. -4, -6
*Answer: I used the quadratic formula to solve and got y=-6,-4. So would my answer be D.? <<< Yes

Solve by factoring: a2 -5a-14 = 0.
a. 5, 14
b. -7, 2
c. 2, 7
d. 7, -2
*Answer: I factored and got a=-2,7 so would my answer be D.?<<< Yes

Solve x2-2x-3>0.
a. (-3, 1)
b. (-1, 3)
c. (- ? , -3) U (1, ? )
d. (- ? , -1) U (3, ? )
*My answer was x<-1 or x>3 so would my answer be B. or D.?<<< D --- Why??


Solve the inequality .<<< Where is the inequality?

a. [-3/2, 3]
b. (- ? , -3] U [3/2, ? )
c. (- ? , -3/2] U [3, ? )
d. [-3, 3/2]
* My answer was -3/2<x<3 so would I choose C.?

Determine the center and radius of the circle (x+2)2+(y-1)2 = 16
a. center: (2, -1); radius: 4
b. center: (2, -1); radius: 16
c. center: (-2, 1); radius: 4
d. center: (-2, 1); radius: 16
* My answer was C. Am I right? <<< Yes
Click to expand...
 
* Thanks so much for your help Subhotosh Khan. Here are the answers to your questions:
Solve x2-2x-3>0.
a. (-3, 1)
b. (-1, 3)
c. (- ? , -3) U (1, ? )
d. (- ? , -1) U (3, ? )
*Did you mean why did I choose choice D. Well I had someone else explain the question to me prior and they told me that since the solution were both x-values, they could not correspond for the (x,y) coordinates. He said that the answer was D but I just could not understand how it could be so thats why I asked the question. Could you tell me why B is the answer and why D is not so I can understand more clearly?

Solve the inequality. |4x-3|?9
a. [-3/2, 3]
b. (- ? , -3] U [3/2, ? )
c. (- ? , -3/2] U [3, ? )
d. [-3, 3/2]
* My answer was -3/2<x<3 so would I choose C.?
Thanx again! :mrgreen:
 
edd fedds said:
* Thanks so much for your help Subhotosh Khan. Here are the answers to your questions:
Solve x2-2x-3>0.
a. (-3, 1)
b. (-1, 3)
c. (- ? , -3) U (1, ? )
d. (- ? , -1) U (3, ? )
*Did you mean why did I choose choice D. Well I had someone else explain the question to me prior and they told me that since the solution were both x-values, they could not correspond for the (x,y) coordinates. He said that the answer was D but I just could not understand how it could be so thats why I asked the question. Could you tell me why B is the answer and why D is not so I can understand more clearly?
Click to expand...

The answer is D.

First, solve \(\displaystyle x^2-2x-3=0\)

\(\displaystyle x=3 \:eek:r\: x=-1\)

Now you have three intervals to check.

[1] x < -1
[2] -1 < x < 3
[3] x > 3

Choose appropriate values to check in your original inequality.

You should find that interval [1] and [2] work just fine.

Therefore, answer D is the correct answer.

edd fedds said:
Solve the inequality. |4x-3|?9
a. [-3/2, 3]
b. (- ? , -3] U [3/2, ? )
c. (- ? , -3/2] U [3, ? )
d. [-3, 3/2]
* My answer was -3/2<x<3 so would I choose C.?
Thanx again! :mrgreen:
Click to expand...

The answer is A.

Actually, your answer should've been \(\displaystyle -\frac{3}{2} \leq x \leq 3\)

The interval is closed. \(\displaystyle \left[-\frac{3}{2}, \: 3\right]\)
 
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