Sadly, I can't see the picture you're looking at as you make that statement.
But I'm pretty sure that when you combine an exterior angle with the interior
angle adjacent to it, you'll wind up with 180°, because they form a linear line.
There actually isn't any such thing as a linear plane.
1. Start with ΔCIJ.
- ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
- the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
- ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.
2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So
m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.
3. Consider ΔCKL.
- ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
- ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
- the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
- ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.
4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So
m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.
5. ΔABC is isosceles, then angles adjacent to the base are congruent:
m∠KBA=m∠JAB → 222°-8x=205°-7x,
7x-8x=205°-222°,
-x=-17°,
x=17°.
Then m∠CAB=m∠CBA=205°-7x=86°.
Answer: 86°.
Answer:
from the top down
y=7
y=4
y=3
y=4
y=7
Step-by-step explanation:
y=(-2)²+3
y=4+3
y=7
y=(-1)²+3
y=1+3
y=4
y=0²+3
y=0+3
y=3
y=1²+3
y=1+3
y=4
y=2²+3
y=4+3
y=7
hope this helps :)
Answer:
she spends 50 hours with her baby and 30 hours on math
Step-by-step explanation:
80/2 = 40
Since baby needs to be 10 more you do 40+10 = 50
and 80 - 50 = 30
Hence the answer
at least i think
