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3 years ago

Please answer this multiple choice question!

Mathematics

2 answers:

Zina [86]3 years ago

7 0

Step-by-step Answer:

There is a theorem in geometry that states that a perpendicular (CQ) to any chord (PR) bisects the chord. This means that PQ=RQ=6cm.

koban [17]3 years ago

4 0

<h3>Answer: Choice B) 6 cm</h3>

==========================

Explanation:

Triangle PQC is congruent to triangle RQC. We can prove this through use of the hypotenuse leg theorem (HL theorem). Note that PC = RC are congruent radii, and also note that CQ = CQ through the reflexive theorem. We have a pair of hypotenuses and a pair of legs for the right triangles.

Then through CPCTC (corresponding parts of congruent triangles are congruent), we know that the pieces PQ and QR are congruent. For now, let's just call them x. They must add to PR which is 12, so,

PQ + QR = PR ..... segment addition postulate

x + x = 12

2x = 12

x = 6 ..... divide both sides by 2

QR = 6

This shows that if you have a radius perpendicular to a chord of a circle, then the radius will bisect this chord. Bisect means to cut in half.

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Answer:

$\dfrac{dh}{dt}=21 \text{m/min}$

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Step-by-step explanation:

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here L is a constant since that won't change as the water is being filled in the trough, however 'b' and 'h' will be changing. The equation has two independent variables and we need to convert this equation so it is only dependent on 'h' (the height of the water).

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The question is asking for the rate of change of height (m/min) hence that can be denoted as: $\frac{dh}{dt}$

Using the chainrule:

$\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}$

the only thing missing in this equation is dh/dV which can be easily obtained by differentiating the volume equation with respect to h

$V = \dfrac{5}{14}h^2L$

$\dfrac{dV}{dh} = \dfrac{5}{7}hL$

reciprocating

$\dfrac{dh}{dV} = \dfrac{7}{5hL}$

plugging everything in the chain rule equation:

$\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}$

$\dfrac{dh}{dt}=\dfrac{7}{5hL}\times 6$

$\dfrac{dh}{dt}=\dfrac{42}{5hL}$

L = 12, and h = 1 (when the water is 1m deep)

$\dfrac{dh}{dt}=\dfrac{42}{5(1)(12)}$

$\dfrac{dh}{dt}=21 \text{m/min}$

the rate of change of height when the water is 1 meter deep is 21 m/min

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