Solution: Let Robot B carry X kilograms per hour.
900=x+3 in X = 600 in X.
Solution: x=60
X=60 is the solution of the original equation.
Type A: 60+30=90 (kg/hour)
A:''''''
4. Solution: Let the speed of armor be 3xkm/h..
3x 6+3 1 = 4x 10.
The solution is x = 3/2.
It is verified that x = 3/2 is the solution of the original equation.
Answer: 3× 3/2 = 9/2 (km/h)
B: 4× 3/2 = 6 (km/h)
A:''''''
Give points!
I may be able to answer your question, because I teach math. It's a pity that I don't have the Eighth Grade Mathematicians' Education Edition at hand, so I'd better post the question, because there are definitely few eighth grade mathematicians.
Mathematics People's Education Edition 18. 1 eighth grade Volume II D C answers the first question of the ninth question, and then proves that the triangle is congruent ∠A=∠B, two heights are equal, and two right angles are equal, so AD=BC is the same as above in the second question, and ∠A=∠B is replaced by AD.
What is the answer to the question 103 1 1 in the second volume of Grade 8? Important process. Using equal area method
Solution: s △ Abd = (1/2) BD * ao =12cm2.
Find the length of AB by Pythagorean theorem, where AB= root sign (AO square +BO square) =5cm.
According to DH⊥AB, s △ Abd = (1/2) ab * DH = (5/2) * DH =12.
DH = 24/5cm
The second volume of the eighth grade Mathematics People's Education Edition, page 37, 9 questions 10, the answer. Urgent! On page 36, the first question of four questions is multiplied by X(X+ 1) and (2X+5) at the same time, and then the score is calculated slowly. The ninth on page 37 is the efficiency of farmers' work based on X.
People's Education Press, eighth grade, Volume II, Mathematics Book, Page 23 14 Answer and process M divided by 1 divided by n+ 1 divided by n-0.5.
The diagonal AC of the diamond of PEP is perpendicular to BD and equally divided with each other, so the area of triangle ABD is equal to AO×BD÷2=(8÷2)×6÷2= 12cm? Is it because of ao again? +Bo? =AB？ , so AB=5cm, and the area of triangle ABD is equal to AB×DH÷2= 12cm? ，DH = 12× 2 ÷ 5 = 4.8cm。
Mathematics P9 1 Page 4, Question 5 Answer 4, ∫ Quadrilateral ABCD is a parallelogram.
∴ AB‖CD，AB=CD，AD‖BC，AD=BC，∠B=∠D
AF = CE
∴AD-AF=BC-CE
∴DF=BE
∴△ABE≌△CDF
∴AE=CF
AEB =∠ computational fluid dynamics
700 BC
∴∠CFD=∠DCB
∴AE‖CF
∴ Quadrilateral AECF is a parallelogram
5.∫E, F, G and H are the midpoint of OA, OB, OC and OD respectively.
∴AE=OE,OF=BF,OG=GC,OH=HD
According to the midline theorem, EF‖AB, EF= 1/2AB.
Similarly, FG‖BC, FG= 1/2BC.
GH‖CD，GH= 1/2CD
EH‖AD，EH= 1/2AD
∵ABCD is a parallelogram.
AB‖CD，AB=CD，AD‖BC，AD=CB
∴EF‖GH,EF=GH,EH‖FG,EH=FG
∴ quadrilateral EFGH is a parallelogram.
The coordinate of point C is (a, 0) in the process of a problem in the second volume of the eighth grade mathematics synchronous tutorial version.
C is on the left side of rule B, and B (a+2,0) A (a+3/2, √ 3/2) (point A is in the first quadrant) gets a = 1/2 C from XY=3 * (√ 3/2) = √ 3.
C is on the right side of rule B, and B (a-2,0) A (a-3/2, √ 3/2) (point A is in the first quadrant) gets a = 7/2 C (7/3) from XY=√3 * (√ 3/2) =√ 3.
C is to the left of rule B, and B (a+2,0) A (a+3/2,-√ 3/2) (point A is in the third quadrant) gets a =-7/2 C from XY=√3 * (√ 3/2) = √ 3.
If point A is in the third quadrant and C is on the right side of rule B, then B (a-2,0) A (a-3/2, -√3/2) is A =- 1 3 from XY=√3 to (a-3/2) * B(a-2 √ 3/2) = √ 3.
The second, third and seventh questions of the eighth grade mathematics exercise 16.3 of People's Education Press should be detailed. The third question:
Solution: Let Robot B carry X kilograms per hour and Robot A carry x+30 kilograms per hour.
900/(x+30) = 600/x Solution: x=60.
A: Type A robots carry 90 kilograms per hour, while Type B robots carry 60 kilograms per hour.
Question 7:
Solution: Let the climbing speed of the second group be x meters per minute. Then the second group is1.2x.
450/x = 450/1.2x+15 solution: x=5 answer: the first group is 6m/min；; The second group is 5m/min.
Let the climbing speed of the second group be x meters per minute. Then the second group is ax.
H/x=h/ax+t solution: x=h/t*( 1- 1/a)
A: the first group is h/t * (a-1) m/min; The second group is h/t*( 1- 1/a) m/min.