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A pile of coal, originally planned to burn 3 tons a day, can burn 96 days. In fact, 2.4 tons a day, how many days can actually burn? (answer in proportion)
It can burn for X days. 3 × 96 = 2.4x & nbsp; & nbsp; X = 3 × 962.4 & nbsp; & nbsp; & nbsp; X = 120 A: it can burn for 120 days
A pile of coal, originally planned to burn 3 tons per day, can burn 72 days. After improving the technology, only 2.4 tons per day can be burned. How many more days can this pile of coal burn than before?
Suppose that this pile of coal can actually burn for X days. From the meaning of the question, we can get: 2.4x = 3 × 72, 2.4x = 216, & nbsp; X = 90; so this pile of coal can burn more than the original: 90-72 = 18 (days). Answer: this pile of coal can burn for 18 days more than the original
There are 105 tons of coal in a factory. It is planned to burn for 15 days. In fact, it burns 3 tons more than planned every day. How many days did it actually burn
Daily = 105 △ 15 = 7 tons
Set the actual burning time for X days
X*(7+3)=105
10X=105
X=105/10
X=10.5
It took 10.5 days
(X-8) + 3x = 42
(X-8) + 3 / 4 x = 42
x-8+(3/4)x=42
(7/4)x=42+8
(7/4)x=50
x=50×(4/7)
x=200/7
2/3x÷4/5=5/16
2X of 3 = 5 of 16 × 4 of 5
2X of 3 = 1 of 4
X = 1 / 4 ／ 2 / 3
X = 3 / 8
3x-2 of 7 = {X-2} * 5 of 7
3x-2 of 7 = (X-2) × 5 of 7
3x-14=5（x-2）
3x-14=5x-10
5x-3x=10-14
2x=-4
x=-2
2.4-2.5 x-4 = 3x of 5
2.4-2.5 x-4 = 5 / 3x, multiply both sides by 10
24-4(x-4)=2*3x
40-4x=6x
2x=40
x=20
3 / 4 (x-1) - 2 / 5 (3x 2) = 1 / 10 - (x-1)
17 out of 7
5 x 10 17-4 3x-7 = - 2
5 x 10 17-4 3x-7 = - 2 (x / 5) + 17 - (3x / 4) - 7 = - 2
(x/5)-(3x/4)=-12
(4x/20)-(15x/20)=-12
(4x-15x)/20=-12
-11x/20=-12
-11x=-240
x=240/11
Is it possible to change y = x-3x + 2 to the square of y = (x-3 / 2) - 1 / 4 and then solve it
Yes,
Shift m units to the right
It is equivalent to y = (x-3 / 2) &# 178; - 1 / 4 to shift m units to the left to get y = (x-3 / 2 + m) &# 178; - 1 / 4
Then the formula of the original equation is compared
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