The second earl of yarborough is reported to have bet at odds of 1000 to 1 that a bridge hand of 13 cards would contain at least one card that is ten or higher. (by ten or higher we mean that a card is either a ten, a jack, a queen, a king, or an ace.) nowadays, we call a hand that has no cards higher than 9 a yarborough. what is the probability that a randomly selected bridge hand is a yarborough?

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

step-by-step explanation: it is d

y= 2x+7

changed to y= 5x +7

as slope is increased from 2 to 5

therefore it would be less steeper

so, option d) is correct

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0