Answer:
71,000 is the nearest thousand
A ruler to measure the length of an object.
. Multiplying the denominator by
gives
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Subtracting this from the numerator gives a remainder of

. Multiplying the denominator by
gives
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and subtracting this from the previous remainder gives a new remainder of

This last remainder is exactly the same as the denominator, so
divides through it exactly and leaves us with 1.
What we showed here is that



and this last expression is the quotient.
To verify this solution, we can simply multiply this by the original denominator:



which matches the original numerator.
Answer:
a) F
b) B, E, D
Step-by-step explanation:
a) The segment with the greatest gradient has the largest change in y-values per unit change in x-values
From the given option, the rate of change of the <em>y </em>to the<em> </em>x-values of B = the gradient = (4 units)/(2 units) = 2
The gradient of F = (-3units)/(1 unit) = -3
The gradient of A = 4/4 = 1
The gradient of C = -2/5
The gradient of D = 2/6 = 1/3
The gradient of E = 3/4
The segment with the greatest gradient is F
b) The steepest segment has the higher gradient
From their calculated we have;
The gradient of segment B = 2 therefore, B is steeper than E that has a gradient of 3/4, and E is steeper than D, as the gradient of D = 1/3
Therefore, we have;
B, E, D.
Let:
Vbu= Volume of the buret
Vbk= Volume of the beaker
A buret initially contains 70.00 millimeters of a solution and a beaker initially contains 20.00 ml of the solution the buret drips solution into the Beaker. each drip contains 0.05 mL of solution, therefore:
x = Number of drips
a = volume of each drip

after how many drips will the volume of the solution in the buret and beaker be equal ? Vbu = Vbk: