John can jog twice as fast as he can walk. he was able to jog the first mile to his grandmas house but then he got tired and walked the remaining 4 miles. if the total trip took 0.75 hours, then what was his average jogging speed

  12 mph

Step-by-step explanation:

The relationship between jogging speed and walking speed means the time it takes to walk 4 miles is the same as the time it takes to jog 8 miles. Then the total travel time (0.75 h) is the time it would take to jog 1+8 = 9 miles. The jogging speed is ...

  (9 mi)(.75 h) = 12 mi/h . . . average jogging speed

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Check

1 mile will take (1 mi)/(12 mi/h) = 1/12 h to jog.

4 miles will take (4 mi)/(6 mi/h) = 4/6 = 8/12 h to walk.

The total travel time is (1/12 +8/12) h = 9/12 h = 3/4 h. (answer checks OK)

Comment on the problem

Olympic race-walking speed is on the order of 7.7 mi/h, so John's walking speed of 6 mi/h should be considered quite a bit faster than normal. The fastest marathon ever run is on the order of a bit more than 12 mi/h, so John's jogging speed is also quite a bit faster than normal. No wonder he got tired.

answer: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer

step-by-step explanation: