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2 years ago

Cheese sticks that were previously priced at "10 for $1" are now "4 for $1". Find each percent change

Mathematics

1 answer:

Naddika [18.5K]2 years ago

5 0

Answer:

The price per cheese stick was increased by 150%.

The amount of cheese sticks per $1.00 has decreased by 60%.

Step-by-step explanation:

The price per cheese stick was $0.10

1.00 (dollar) divided into 10 (cheese sticks) gives you 0.1, which means each cheese stick costs $0.10

The current price per cheese stick is $0.25

1.00 (dollar) divided into 4 (cheese sticks) gives you 0.25, which means each cheese stick costs $0.25

That means the price per cheese stick was raised by 150%.

That also means the amount of cheese sticks you could buy for $1.00 has decreased by 60%.

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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

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Given: circle k(O), m LM = 164° m WK = 68°, m∠MLK = 65° Find: m∠LMW

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3 years ago

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