Side CB is the same as AD.
Side AB is the same as CD.
<CAD is 15 degrees
<ACD is 24 degrees
now use a^2 +b^2=c^2 to find the last angle
Since we are solving for b, you need to get it by itself. So you would divide both sides by 3a. That would leave b= -7/3a.
Hope this helps!!
Answer:
x = 12
m(QS) = 52°
m(PD) = 152°
Step-by-step explanation:
Recall: Angle formed by two secants outside a circle = ½(the difference of the intercepted arcs)
Thus:
m<R = ½[m(PD) - m(QS)]
50° = ½[(12x + 8) - (4x + 4)] => substitution
Solve for x
Multiply both sides by 2
2*50 = (12x + 8) - (4x + 4)
100 = (12x + 8) - (4x + 4)
100 = 12x + 8 - 4x - 4 (distributive property)
Add like terms
100 = 8x + 4
100 - 4 = 8x
96 = 8x
96/8 = x
12 = x
x = 12
✔️m(QS) = 4x + 4 = 4(12) + 4 = 52°
✔️m(PD) = 12x + 8 = 12(12) + 8 = 152°
Answer:
x < -11
Step-by-step explanation:
Let x be the number
- 2x - 2 < -24
- Add 2 to each side, so it now looks like this: 2x < -22
- Divide each side by 2 to cancel out the 2 next to x. It should now look like this: x < -11
I hope this helps!
