Answer:
idk what exactly this is asking buut here is my best answer
Step-by-step explanation:
1+1=2(3+5=-1)
since 6 is the possible outcoe .idk what its exacly askign so if u can help with tht that wuld be grest .
Need 1 more brainliest
Answer:
yes
Step-by-step explanation:
Use Pythagorean Theorem:
10² + 49.5² = 50.5² ??
100 + 2450.25 = 2550.25
2550.25 = 2500.25
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:
20 km
Step-by-step explanation:
We will imagine a triangle. The hypotenuse of the triangle is the distance between you and the plane. 16 kilometres is the aerial space between you two, in other words its the base of the triangle. 12 km is the length of the third leg of the triangle. We will apply the pythagorean theorem to find the length of the hypotenuse and the distance between you and the plane.
a^2 + b^2 = c^2
12^2 + 16^2 = c^2
144 + 256 = c^2
400 = c^2
square root of 400 is c which equals 20 km.
