Which pair of expressions is equivalent
1 answer:
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Let
x-----------> first <span>odd integer
x+2--------> second consecutive odd integer
x+4-------> third consecutive odd integer
we know that
(x+4)</span>²=15+x²+(x+2)²-------> x²+8x+16=15+x²+x²+4x+4
x²+8x+16=19+2x²+4x-------> x²-4x+3
x²-4x+3=0
using a graph tool----------> to calculate the quadratic equation
see the attached figure
the solution is
x=1
x=3
the answer is
the first odd integer x is 1
the second consecutive odd integer x+2 is 3
the third consecutive odd integer x+4 is 5
x-----------> first <span>odd integer
x+2--------> second consecutive odd integer
x+4-------> third consecutive odd integer
we know that
(x+4)</span>²=15+x²+(x+2)²-------> x²+8x+16=15+x²+x²+4x+4
x²+8x+16=19+2x²+4x-------> x²-4x+3
x²-4x+3=0
using a graph tool----------> to calculate the quadratic equation
see the attached figure
the solution is
x=1
x=3
the answer is
the first odd integer x is 1
the second consecutive odd integer x+2 is 3
the third consecutive odd integer x+4 is 5
Answer:
Step-by-step explanation:
In a deck of cart, we have:
a = 4 (aces)
t = 4 (three)
j = 4 (jacks)
And the total number of cards in the deck is
n = 52
So, the probability of drawing an ace as first cart is:
At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is
Therefore, the probability of drawing a three at the 2nd draw is
Then, at the third draw, the previous 2 cards are not replaced, so there are now
cards in the deck. So, the probability of drawing a jack is
Therefore, the total probability of drawing an ace, a three and then a jack is: