A single card is drawn from a deck of 52 cards, find the following
1 answer:
You might be interested in
Judy's distance from the intersection: d
<span>When Jackie is 1 mile farther from the intersection than Judy:
</span>Jakie's distance from the intersection: d+1
T<span>he distance between them is 2 miles more than Judy’s distance from the intersection: d+2
</span>How far is Jackie from the intersection?
d+1=?
Using the Pythagoras Theorem:
c^2=a^2+b^2; a=d, b=d+1, c=d+2
(d+2)^2=(d)^2+(d+1)^2
(d)^2+2(d)(2)+(2)^2=d^2+(d)^2+2(d)(1)+(1)^2
d^2+4d+4=d^2+d^2+2d+1
d^2+4d+4=2d^2+2d+1
d^2+4d+4-d^2-4d-4=2d^2+2d+1-d^2-4d-4
0=d^2-2d-3
d^2-2d-3=0
Factoring:
(d-3)(d+1)=0
d+1=0→d+1-1=0-1→d=-1<0 (negative): No possible. The distance can't be negative.
d-3=0→d-3+3=0+3→d=3 miles>0 (positive): Ok
Jakie's distance from the intersection=d+1=3+1→d+1=4 miles
Answer: <span>Jackie is 4 miles far from the intersection
</span>
Please, see the attached file.
Thanks
<span>When Jackie is 1 mile farther from the intersection than Judy:
</span>Jakie's distance from the intersection: d+1
T<span>he distance between them is 2 miles more than Judy’s distance from the intersection: d+2
</span>How far is Jackie from the intersection?
d+1=?
Using the Pythagoras Theorem:
c^2=a^2+b^2; a=d, b=d+1, c=d+2
(d+2)^2=(d)^2+(d+1)^2
(d)^2+2(d)(2)+(2)^2=d^2+(d)^2+2(d)(1)+(1)^2
d^2+4d+4=d^2+d^2+2d+1
d^2+4d+4=2d^2+2d+1
d^2+4d+4-d^2-4d-4=2d^2+2d+1-d^2-4d-4
0=d^2-2d-3
d^2-2d-3=0
Factoring:
(d-3)(d+1)=0
d+1=0→d+1-1=0-1→d=-1<0 (negative): No possible. The distance can't be negative.
d-3=0→d-3+3=0+3→d=3 miles>0 (positive): Ok
Jakie's distance from the intersection=d+1=3+1→d+1=4 miles
Answer: <span>Jackie is 4 miles far from the intersection
</span>
Please, see the attached file.
Thanks