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xz_007 [3.2K]
2 years ago
11

2) What is the length of BD and BC in the diagram below?

Mathematics
1 answer:
babunello [35]2 years ago
4 0

Answer:

BC is 10 unit

BD is 15 units

Step-by-step explanation:

Here, given the end points of the line segments, we want to calculate BD and DC

Mathematically, the distance between two points can be calculated using the formula;

D = √(x2-x1)^2 + (y2-y1)^2

For BD, we have the following

D = √(14-5)^2 + (14-2)^2

D = √(9)^2 + 12^2

D = √81 + 144

D = √(225)

Distance BD is 15 units

For BC, we have the following

D = √(11-5)^2 + (10-2)^2

D = √(6)^2 + 8^2

D = √100

D = 10

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The diagram on the right shows the graph of a quadratic function f(x)=kx^2+6x+h.Point A(3,14) is the maximum point of the graph
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A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4
Vladimir79 [104]

Answer:

The time interval when V_Q(t) \geq 60  is at  1.866 \leq t \leq 3.519

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Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

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V_Q(t) \geq 60    between   0 \leq t \leq 4

The schematic free body graphical representation of the above illustration was attached in the file below and the point when V_Q(t) \geq 60  is at 4 is obtained in the parabolic curve.

So, V_Q(t) \geq 60  is at  1.866 \leq t \leq 3.519

Taking the integral of the time interval in order to determine the distance; we have:

distance = \int\limits^{3.519}_{1.866} {V_Q(t)} \, dt

= \int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt

= By using the Scientific calculator notation;

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3 years ago
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2 years ago
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