2/5(d−10)−23(d+6)25(d−10)−23(d+6) is
1 answer:
Simplify 
(d - 10) to 
:
\frac{2(d - 10)}{5} [/tex] - 23(d + 6) x 25(d - 10) - 23(d + 6)
Simplify 23(d + 6) x 25(d - 10) to 575(d + 6)(d - 10) :
\frac{2(d - 10)}{5} [/tex] - 575(d + 6)(d - 10) - 23(d + 6)
Then Expand :
\frac{2(d - 10)}{5} [/tex] - 575
+ 5750d - 3450d + 34500 - 23d - 138
Now Collect Like Terms :
\frac{2(d - 10)}{5} [/tex] - 575 + (5750d - 3450d + 34500 - 23d) + (34500 - 138)
Answer :
\frac{2(d - 10)}{5} [/tex] - 575 + 2277d + 34362
\frac{2(d - 10)}{5} [/tex] - 23(d + 6) x 25(d - 10) - 23(d + 6)
Simplify 23(d + 6) x 25(d - 10) to 575(d + 6)(d - 10) :
\frac{2(d - 10)}{5} [/tex] - 575(d + 6)(d - 10) - 23(d + 6)
Then Expand :
\frac{2(d - 10)}{5} [/tex] - 575
Now Collect Like Terms :
\frac{2(d - 10)}{5} [/tex] - 575
Answer :
\frac{2(d - 10)}{5} [/tex] - 575
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Answer:
d = 15 miles
Step-by-step explanation:
To get to his parents’ house, Ruben would have to drive due north 12 miles.
To get to his grandparents’ house, he would have to drive due east 9 miles.
It is required to find the distance between the parents’ and grandparents’ houses.
First distance is 12 miles and second distance is 9 miles.
So,
So, the distance between the parents’ and grandparents’ houses is 15 miles.