Given:

RS tangent to circle k(O) at R,

TU tangent to circle k(O) at T,

SU tangent to circle k(O) at V,

OS ⊥ OU.

Prove: RS ∥ TU

use substitution to solve each system of equations. y = x + 5 3x + y = 25 62/87,21 y = x + 5 3x + y = 25 substitute x + 5 for y in the second equation. substitute the solution for x into either equation to find y. the solution is (5, 10)

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step-by-step explanation:

d

step-by-step explanation:

given

y + 3 = - 4(x - 2) ← distribute the parenthesis

y + 3 = - 4x + 8 ( subtract 3 from both sides )

y = - 4x + 5 → d

72 is the answer to this problem

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560