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mixer [17]
3 years ago
8

What is the area of triangle FGH? Round your answer to the nearest tenth of a square centimeter.

Mathematics
1 answer:
Misha Larkins [42]3 years ago
8 0

Answer:

Step-by-step explanation:

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y = 1x + -1 or y = x - 1

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$9.50

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equal to 4

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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

The distance is 106.109 m

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.

We are also to that :

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;

distance = 106.109 m

4 0
3 years ago
Question 71
yuradex [85]

Answer:

given

p=q^3

p=40

______

q=(p)^1/3

q=(40)^1/3-->q=2(5)^1/3

Now

the value of half of q , (2(5)^1/3)÷(2) ,is 5^1/3.

finally the value of p gonna be

p=q^3

p=(5^1/3) ^ 3

p=5

3 0
3 years ago
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