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lawyer [7]
4 years ago
15

Michael started a savings account with $300. After 4 weeks, he had $350, abd after 9 weeks, he had $400. What is the rate of cha

nge?
Mathematics
1 answer:
Colt1911 [192]4 years ago
6 0

Answer:


Step-by-step explanation:

The initial amount was $300 and the final amount $350.  This $50 change occurred in 4 weeks.  Thus, the rate of change was

  $50

----------- = $12.50/week for the first four weeks

 4 wks


For the first nine weeks, the rate of change was


$400-$300          $100

------------------- = ----------------- = $0.11/week.

       9 wks               9 wks


Note that this second result makes little sense.  Ensure that you have copied this problem down correctly.

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If a snowball melts so that its surface area decreases at a rate of 3 cm2/min, find the rate (in cm/min) at which the diameter d
grin007 [14]

Answer:

The diameter decreases at a rate of 0.053 cm/min.

Step-by-step explanation:

Surface area of an snowball

The surface area of an snowball has the following equation:

S_{a} = \pi d^2

In which d is the diameter.

Implicit differentiation:

To solve this question, we differentiate the equation for the surface area implictly, in function of t. So

\frac{dS_{a}}{dt} = 2d\pi\frac{dd}{dt}

Surface area decreases at a rate of 3 cm2/min

This means that \frac{dS_{a}}{dt} = -3

Tind the rate (in cm/min) at which the diameter decreases when the diameter is 9 cm.

This is \frac{dd}{dt} when d = 9. So

\frac{dS_{a}}{dt} = 2d\pi\frac{dd}{dt}

-3 = 2*9\pi\frac{dd}{dt}

\frac{dd}{dt} = -\frac{3}{18\pi}

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The diameter decreases at a rate of 0.053 cm/min.

7 0
3 years ago
Describe the rotation of J
Serga [27]

Answer:

The coordinates of J' when rotating by 90° counterclockwise will be: J'(3, 1)

The coordinates of J' when rotating by 90° clockwise will be: J'(-3, -1)

Step-by-step explanation:

Square JKLM with vertices

  • J(1, -3)
  • K(5, 0)
  • L(8, -4)
  • M(4, -7)

We have to determine the answer for the image J' of the point (1, -3) when we rotate the point by 90° counterclockwise, we need to switch x and y, make y negative.

In other words, the rule to rotate a point by 90° counterclockwise.

P(x, y) → P'(-y, x)

As we are given that J(1, -3), so the coordinates of J' will be:

J(1, -3) → J'(3, 1)

Therefore, the coordinates of J' when rotating by 90° counterclockwise will be:  J'(3, 1).

When the point is rotated by 90° clockwise, we need to switch x and y, make x negative.

In other words, the rule to rotate a point by 90° clockwise.

P(x, y) → P'(y, -x)

As we are given that J(1, -3), so the coordinates of J' will be:

J(1, -3) → J'(-3, -1)

Therefore, the coordinates of J' when rotating by 90° clockwise will be:  J'(-3, -1)

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