Question 1A.
The given expression, particularly its last term, may be interpreted in different ways. To avoid the confusion it is rewritten in the following form.
6ag^2 - 4a + (5g^2)a
= 6ag^2 - 4a + 5ag^2 ... [As ag^2 is equal to (g^2)a]
= 11ag^2 - 4a
= a(11g^2 - 4)
Question 1B.
2x(5x-8)-(6x-3)
= 10x^2 - 16x - 6x + 3
= 10x^2 - 22x + 3
Question 2
Volume of drum in litres is given by the formula:
Volume = [(Pi)*(Radius)^2*(Height)]/1000
when both radius and height are measured in cm.
In the question the value of radius has not been given, whereas that of the height of drum is given which has no bearing on the problem. Assuming height of drum mentioned is actually meant to be radius, the given values are:
Radius = 1m = 100cm
Volume = 25 litre
Substituting these values in the equation for volume:
25 = [3.14159*100^2*(Height)]/1000 = 31.4159*Height
Therefor:
Height = 25/31.4159 = 0.79 = 0.8 cm (rounded off to nearest first decimal place)
Thus if water contains less than 25 litre of water the depth of water column will be less than 0.8 cm.
Question 3.
Let the speed on the dirt road be equal to x.
Therefor speed on the high is equal to (x + 30)
Distance travelled on dirt road
= (Speed on dirt road)*(Time on dirt road) = x*1 = x
Distance travelled on highway
= (Speed on highway)*(Time on highway) = (x + 30)*3 = 3x + 90
Total distance travelled on dirt road plus on highway
= x + 3x + 90 = 4x + 90
This is given to be equal to 330 km
Therefor
4x + 90 = 330
4x = 330 - 90 = 240
and:
x = 240/4 = 60 km/h = Speed on the dirt road.
Speed on highway = x + 30 = 60 + 30 = 90 km/h
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