Answer:
35, 48 63 80 99
Step-by-step explanation:
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Answer:
2. Like terms: 5a 4. Lt: 4y, and y
coefficient: 2,-7 coefficient: 4, -3
Constant: 2,-7 constant: 4
3. like terms: 3h, 2h, and 6h can you follow these and do the remaining??
if no, I'll help u
coefficient: 3,2,6
constant: 9
Answer:
74
Step-by-step explanation:
Say that arc JL going through M is arc E and JL going the other way is arc D
For the angle formed by two tangents, K=(1/2)(E-D)
64=E-D
Furthermore, angle K and central angle JCL (facing toward K) are supplementary, so 180-K=JCL=180-32=148
Thus, as the angles around angle C add up to 360, angle JCL (facing toward M) is 360-148=32+180=212
E is then 212
64=212-D
212-64=D=148
Thus, as JML is an inscribed angle, M=1/2(D)=1/2(148)=74
For number 7.
Supplementary angles sum to 180.
2x+4x=180
6x=180
x=30
The two angles are 2x and 4x or 60 and 120
Answer:
2 - 2y^2 - x/y
Step-by-step explanation:
2x^2/x^2 = 2
and
x^2/xy = x/y
therefore
2x^2/x^2 - y^2 - x^2/xy - y^2
= 2 - y^2 - x/y - y^2
= 2 - y^2 - y^2 - x/y
= 2 - 2y^2 - x/y
