Answer:
5 cm
Step-by-step explanation:
If AB is tangent to the circle k(O), then radius OB is perpendicular to segment AB.
If BC is tangent to the circle k(O), then radius OC is perpendicular to segment AC.
Consider two right triangles ABO and ACO. In these triangles:
- AO is common hypotenuse;
- ∠OBA=∠OCA=90°, because AB⊥OB, AC⊥OC;
- OB=OC as radii of the circle k(O).
By HL theorem, triangles ABO and ACO are congruent. Then
- ∠OAB=∠OAC=30°;
- AC=AB=5 cm.
Hence, ∠BAC=∠OAB+∠OAC=30°+30°=60°.
Consider triangle ABC, this triangle is isosceles triangle. In isosceles triangles angles adjacent to the base are congruent, thus
∠CBA=∠BCA=1/2(180°-60°)=60°.
Therefore, triangle ABC is an equilateral triangle, so BC=AB=AC=5 cm.
2.7 is 27/10 and 3.6 is 18/5
Answer:
173 days
Step-by-step explanation:
The formula for substantial presence test is;
SBT = (Total of number of days present in the current tax year) + (1/3)(number of days in the year that was before the tax year) + (1/6)(number of days in the year that was two years before the tax year)
From the question, present tax year is 2019 and number of days is 96 days.
Year before tax year is 2018 and number of days is 198 days
2 years before tax year is 2017 and number of days is 66 days.
Thus;
SBT = 96 + ((1/3)198) + ((1/6)66)
SBT = 173 days
Answer:
- see the attachment for a graph
- domain: all real numbers
- range: -6 and y>-5
Step-by-step explanation:
The graph below was created by a graphing utility.
It shows no horizontal gaps: f(x) is defined for all values of x, so the domain is all real numbers.
It shows a vertical gap between y=-6 and y>-5. So, the range is in two parts:
{-6} ∪ {y > -5}
Answer: 1 SSS
Explanation: I hope this helps you
