英文中四數學 - Ans.fyi 參考答案

英文中四數學

2011-07-21 6:26 pm

Please to solve the following questions:

1.384 sweets are packed evenly into bags. If the number of sweets in each bag
is 8 less than the total number of bags packed, how many sweets are there in eanch bag?

2.The area of a rectangular swimming pool is 500m^2. If the length of the pool is decreased by 5m, a square pool will be formed. Find the original lengthe and
width of the swimming pool.

3.A beanbag initially at rest is dropped from the top of a building. The distance
(h m) that it falls after t seconds is given by h=5t^2. Find the time taken
by the beanbag to fall a distance of 45 m.

4. The height of a right-angled triangle is 7 cm longer than the base. If the
perimeter of the right-angled triangle is 30cm, find the height of the
triangle.

5. The price ($P) of a circular medal with radius x cm is given by
P=2X^2 - 18X + 55. Can the price of a circular medal be $11?
Explain your answer.

6. Two numbers differ by 9. Is it possible that the sum of the squares of the two numbers is 39? Explain your answer.

Thank!

回答 (1)

2011-07-22 7:17 am

✔ 最佳答案

1.
let the number of sweets in each bag be y
the total number of bags packed=y+8 number of bags*number of sweets in each bag=total number of sweets
(y+8)(y)=384
y^2+8y-384=0
(y-16)(y+24)
y=16 or y=-24 (rejected)the number of sweets in each bag is 162.
let the original length be y m
Since the length of the pool is decreased by 5m,
a square pool will be formed
the width=y-5 m y*(y-5)=500
y^2-5y-500=0
(y-25)(y+20)=0
y=25 or y=-20 (rejected)the original length is 25m3.
h=5(t^2)
when h=45,
45=5(t^2)
9=t^2
t^2-9=0
t^2-3^2=0
(t-3)(t+3)=0
t=3 or t=-3 (rejected)
the time taken is 3 seconds 4.
let the height of the triangle be h cm
its base=h-7 cm Since the perimeter=30cm,
the length of the hypotenuse
(or the remaining side)
=30-h-(h-7) cm
=37-2h cmBy Pythagoras' Theorem,
(37-2h)^2=(h)^2+(h-7)^2
1369-148h+4(h^2)=h^2+h^2-14h+49
2(h^2)-134h+1320=0
h^2-67h+660=0
(h-55)(h-12)=0
h=55 (rejected) or h=12the height of the triangle is 12cm5.
when P=11,
we have
11=2(x^2)-18x+55
2(x^2)-18x+44=0
x^2-9x+22=0
(x-9/2)^2-81/4+22=0
(x-9/2)^2+7/4=0
Contradiction occurs!
(since (x-9/2)^2+7/4>0+7/4>0 ) So, the price of a circular medal cannot be $116.
let the smaller number be y
the another number=y+9 y^2+(y+9)^2=39
y^2+y^2+18y+81=39
2(y^2)+18y+42=0
y^2+9y+21=0
(y+9/2)^2-81/4+21=0
(y+9/2)^2+3/4=0
Contradiction occurs!
(since (y+9/2)^2+3/4>0+3/4>0 )So,it is not possible that
the sum of the squares of the two numbers is 39

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