We know that
If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment. (Intersecting Secant-Tangent Theorem)
so
ST²=RT*QT
RT=7 in
QT=23+7-----> 30 in
ST²=7*30-----> 210
ST=√210-----> 14.49 in
the answer is
RT=14.49 in
Answer:
So x = 20
Step-by-step explanation:
AB + BC = AC sum of a segment is sum of its pieces
AB + AB = AC B is the midpoint, so AB = BC
2x + 2x = 3x + 20 putting in for AB (twice!) and AC
4x = 3x + 20 collect like terms
x = 20 subtract 3x on both sides.
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Answer:
12, 17, 24
Step-by-step explanation:
To find the first 3 terms, substitute n = 1, 2, 3 into the rule, that is
T₁ = 1² + 2(1) + 9 = 1 + 2 + 9 = 12
T₂ = 2² + 2(2) + 9 = 4 + 4 + 9 = 17
T₃ = 3² + 2(3) + 9 = 9 + 6 + 9 = 24
Answer:
dy/dx = d/dx (sin3x.cos2x)
dy/dx = cos3x . 3 . -sin2x . 2 (Using chain rule)
dy/dx = -6cos3xsin2x
If your teacher is asking "which of the following can be used to prove the triangles congruent?" then I agree with your statement that it's "none of the above". We simply don't have enough information to determine if the triangles are congruent or not. If we wanted to use SAS, then we'd have to know if EB = BD was true. If we wanted ASA, then we'd have to know that A = C. If we wanted AAS, then we'd have to know that E = D.
In short, you have the correct answer. Nice work.
