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slamgirl [31]
2 years ago
6

Max has a Starbucks' gift card for $50. HIs favorite drink is the S'mores Frappuccino. Each drink costs $4.25. How many frappucc

inos can Max purchase with his gift card? Write an inequality to represent the number of frappuccinos Max can purchase. How many frappuccinos can Max purchase with his gift card?
Mathematics
1 answer:
hram777 [196]2 years ago
5 0

Answer:

4.25x ≤ 50

Number of drink = 11

Step-by-step explanation:

Given:

Amount of gift card = $50

Each drink cost = $4.25

Find:

Number of drink

Computation:

Assume;

Number of drink = x

So,

4.25x ≤ 50

x ≤ 11.76

so,

Number of drink = 11

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horsena [70]

9514 1404 393

Answer:

  • resultant force: 93.946∠-10.62° N
  • line of action: 17.314x +92.337y = 809.433

Step-by-step explanation:

We can use the notation a∠b to represent the (x, y) components (a·cos(b), a·sin(b)), where angle b is measured CCW from the +x direction. If we label the forces a, b, c, d clockwise from A, then we have ...

  a = 80∠0° = (80, 0)

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  c = 90∠45° = (63.640, 63.640)

  d = 150∠-110° = (-51.303, -140.954)

__

If we label point A the origin, then the clockwise torque on point A is the sum of products of the force x-component and its y location, and its y-component and the negative of its x location.

  T = (0, 0)·(80, 0) +(3, 0)·(0, 60) +(3, -8)·(63.640, 63.640) +(0, -8)·(-51.303, -140.954)

  T = 809.433 . . . . n·m, the CW torque on point A

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The sum of forces is ...

  F = a +b +c +d = (92.337, -17.314) = 93.946∠-10.62° . . . N

__

This force, applied to the point of application, must generate the same torque as the given forces. That is ...

  F·(y, -x) = 809.433

Then the equation of the line of action is ...

  17.314x +92.337y = 809.433 . . . . . x and y in meters measured from A

Any point (x, y) on this line will serve as a point of application of the force. Unfortunately, this line of action does not pass through the rectangular plate. The attachment shows the point (D) on the line of action that is closest to point A.

_____

<em>Additional comment</em>

The resultant force could be decomposed into two forces acting <em>on the rectangular plate</em>. One could be of much larger magnitude, operating at the corner opposite point A. This force would provide the necessary torque. Another would be acting on point A, providing no torque, but with components such that the resultant has the correct magnitude and direction.

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