Answer:
60
Step-by-step explanation:
Tangents to a circle from a common external point are congruent, thus
JA = JB = 6
AL = KC = 11
CK = KB = 13
Thus
perimeter = 2(6) + 2(11) + 2(13) = 12 + 22 + 26 = 60
0x39=n
anything times zero = zero so that mean n = 0
<h3>
Answer: b = 4 and c = 7.</h3>
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Explanation:
Comparing y = x^2+bx+c to y = ax^2+bx+c, we see that a = 1.
The vertex given is (-2,3). In general, the vertex is (h,k). So h = -2 and k = 3.
Plug those three values into the vertex form below
y = a(x-h)^2 + k
y = 1(x-(-2))^2 + 3
y = (x+2)^2 + 3
Then expand everything out and simplify
y = x^2+4x+4 + 3
y = x^2+4x+7
We see that b = 4 and c = 7.
Answer:
3
Step-by-step explanation:
To solve this problem, you need to use the Phythagorean theorem c^2 is equal to the sum of a^2 and b^2. If a less than or equal to b and b is less than or equal to c or a^2 plus b^2 is greater than c^2 then it is an acute angle.
