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2 years ago

3 A farmer has 580 square bales of hay in his shed with total mass 9850 kg.

Mathematics

1 answer:

rusak2 [61]2 years ago

4 0

Answer:

a) 580 - 16n

b) 9850 - 271.72n

Step-by-step explanation:

a) starting number of bales = 580

each day, the farmer feeds out 2 bales of hay to each yard. he has 8 yards, so he feeds out 2 bales 8 times for a total of 2 * 8 = 16 bales. Therefore, he loses 16 bales each day

After the first day, he has 580-16 = 564 bales. After the second, he has 564-16 = 548 bales, and so on. He loses 16 bales n times for n days, and as a result, his final amount of bales is

original amount - amount lost = 580 - 16n

First, we can calculate how much a bale weighs.

average = sum/count = 9850 kg / 580 ≈ 17 kg

Therefore, as he loses 16 bales of hay, he loses 17 kg 16 times, or approximately 271.72 kg of hay every day. For n days, he loses 271.72n kg. This brings his final amount to be

starting - amount lost = 9850 - 271.72n

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2 years ago

XY is a diameter of a circle and Z is a point on the circle such that ZY=6. If the area of the triangle XYZ is 18 square root 3

nataly862011 [7]

<h2>Answer:</h2>

4π

<h2>Step-by-step explanation:</h2>

As shown in the diagram, triangle XYZ is a right triangle. Therefore, its area (A) is given by:

A = $\frac{1}{2}$ x b x h -------------(i)

Where;

A = 18 $\sqrt{3}$

b = XZ = base of the triangle

h = YZ = height of the triangle = 6

<em>Substitute these values into equation(i) and solve as follows:</em>

18 $\sqrt{3}$ = $\frac{1}{2}$ x b x 6

18 $\sqrt{3}$ = 3b

<em>Divide through by 3</em>

6 $\sqrt{3}$ = b

Therefore, b = XZ = 6 $\sqrt{3}$

<em>Now, assume that the circle is centered at O;</em>

Triangle XOZ is isosceles, therefore the following are true;

(i) |OZ| = |OX|

(ii) XZO = ZXO = 30°

(iii) XOZ + XZO + ZXO = 180° [sum of angles in a triangle]

=> XOZ + 30° + 30° = 180°

=> XOZ + 60° = 180°

=> XOZ = 180° - 60°

=> XOZ = 120°

Therefore we can calculate the radius |OZ| of the circle using sine rule as follows;

$\frac{sin|XOZ|}{XZ} = \frac{sin|ZXO|}{OZ}$

$\frac{sin120}{6\sqrt{3} } = \frac{sin 30}{OZ}$

$\frac{\sqrt{3} /2}{6\sqrt{3} } = \frac{1/2}{|OZ|}$

$\frac{1}{12} = \frac{1}{2|OZ|}$

$\frac{1}{6} = \frac{1}{|OZ|}$

|OZ| = 6

The radius of the circle is therefore 6.

<em>Now, let's calculate the length of the arc XZ</em>

The length(L) of an arc is given by;

L = θ / 360 x 2 π r ------------------(ii)

Where;

θ = angle subtended by the arc at the center.

r = radius of the circle.

In our case,

θ = ZOX = 120°

r = |OZ| = 6

Substitute these values into equation (ii) as follows;

L = 120/360 x 2π x 6

L = 4π

Therefore the length of the arc XZ is 4π

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