Answer:
WU = (14√13)/13 ≈ 6.6564
Step-by-step explanation:
Call the incenter of ∆KWU point A. Call the center of circle ω2 point B.
Then ∠KWA has half the measure of arc WA. ∠AWU is congruent to ∠KWA, so also has half the arc measure. That is, ∠KWU has the same measure as arc WA and ∠KBW.
KB is a perpendicular bisector of chord WU, so ∆KWB is a right triangle, of which WU is twice the altitude to base KB.
The length of KB can be found several ways. One of them is to use the Pythagorean theorem:
KB² = KW² +WB² = 4² +6² = 52
KB = √52 = 2√13
The area of triangle KWB is ...
area ∆KWB = (1/2)KW·WB = (1/2)(4)(6) = 12 . . . . square units
Using KB as the base in the area calculation, we have ...
area ∆KWB = (1/2)(KB)(WU/2)
12 = KB·WU/4
WU = 48/KB = 48/(2√13) = 24/√13
WU = (24√13)/13 ≈ 6.6564
Answer:
The vertex of the function is (5,6).
Step-by-step explanation:
The equation is already written in vertex form. Given a quadratic function with a vertex of (h,k), the equation for the graph is y=a(x-h)^2+k.
We can extract h and k from the provided equation and get the point (5,6)
It is the type of triangle, in which all the three sides are equal to each other.
Hope this helps!
Answer:
412
Step-by-step explanation:
You have to follow PEMDAS and solve the problem carefully.


Answer: <em><u>4.</u></em> This is because 1 and 1/5 = 1.2 and 3/10 = 0.3 and 1.2 divided by 0.3 = 4. That's because 1.2 can be split into 4 equal groups and each group has 0.3.