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const2013 [10]
3 years ago
15

A rectangle has an area of 216.24 cm2 and a height of 15.9 cm. How long is the base of the rectangle ?

Mathematics
1 answer:
Helga [31]3 years ago
5 0
The formula for the area of a rectangle is A = bh.
Substitute 216.24 for A (area) and 15.9 for h (height).

A = bh
216.24 = b(15.9)
÷ 15.9      ÷ 15.9            Divide 15.9 on both sides.
13.6 = b

So, the base of the rectangle is 13.6 cm.
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0.5, 0.25, 0.125

Step-by-step explanation:

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If a car has tires with a diameter of 28 inches and is traveling at 55 mph, how fast are its tires spinning, in rpm^ prime s (re
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Answer:

The tires are spinning at 660.49 revolutions per minute.

Step-by-step explanation:

The speed of the tires (v) is the same that the speed of the car, so to find the angular velocity of the tires we need to use the equation:

\omega = \frac{v}{r}

Where:

r: is the radius of the tires = d/2 = 28 inches/2 = 14 inches

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Therefore, the tires are spinning at 660.49 revolutions per minute.

   

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3 years ago
The next model of a sports car will cost 4.4% more that the current model. The current model costs $48,000, How much will the pr
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Answer:

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Step-by-step explanation:

1st way.  $48,000 x 1.044 = $50,112.  1.044 is the same thing as saying 104.4% of the total price in decimal form, which is what the new price would be when adding 4.4% to the 100% of the original price.  To get the amount of increase just find the difference in the two.  $50,112-$48,000 = $2,112

2nd way.  $48,000 x .044.  =  $2,112 .044 is 4.4% in decimal form. So you found the increase here.  Add the increase to the original price.  $48,000 + $2,112 to get the new price of $50,112.

7 0
2 years ago
Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre
Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

(c) Inflection point: (0.56,-0.06)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = \frac{d}{dx}[x^{4}ln(x)]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = 4x^{3}ln(x) + x_{4}.\frac{1}{x}

f'(x) = 4x^{3}ln(x) + x^{3}

f'(x) = x^{3}[4ln(x) + 1]

Now, find the critical points: f'(x) = 0

x^{3}[4ln(x) + 1] = 0

x^{3} = 0

x = 0

and

4ln(x) + 1 = 0

ln(x) = \frac{-1}{4}

x = e^{\frac{-1}{4} }

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval                 x-value                      f'(x)                       result

0<x<0.78                 0.5                 f'(0.5) = -0.22            decreasing

x>0.78                       1                         f'(1) = 1                  increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

  • Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
  • After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = x^{4}ln(x)

f(0.78) = 0.78^{4}ln(0.78)

f(0.78) = - 0.092

The point of <u>minimum</u> is (0.78, - 0.092)

(c) To determine the inflection point (IP), calculate the second derivative of the function and solve for x:

f"(x) = \frac{d^{2}}{dx^{2}} [x^{3}[4ln(x) + 1]]

f"(x) = 3x^{2}[4ln(x) + 1] + 4x^{2}

f"(x) = x^{2}[12ln(x) + 7]

x^{2}[12ln(x) + 7] = 0

x^{2} = 0\\x = 0

and

12ln(x) + 7 = 0\\ln(x) = \frac{-7}{12} \\x = e^{\frac{-7}{12} }\\x = 0.56

Substituing x in the function:

f(x) = x^{4}ln(x)

f(0.56) = 0.56^{4} ln(0.56)

f(0.56) = - 0.06

The <u>inflection point</u> will be: (0.56, - 0.06)

In a function, the concave is down when f"(x) < 0 and up when f"(x) > 0, adn knowing that the critical points for that derivative are 0 and 0.56:

f"(x) =  x^{2}[12ln(x) + 7]

f"(0.1) = 0.1^{2}[12ln(0.1)+7]

f"(0.1) = - 0.21, i.e. <u>Concave</u> is <u>DOWN.</u>

f"(0.7) = 0.7^{2}[12ln(0.7)+7]

f"(0.7) = + 1.33, i.e. <u>Concave</u> is <u>UP.</u>

4 0
3 years ago
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