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shutvik [7]
3 years ago
7

Find the product of 21 and (6.3 · 10^-4).0.1323 · 10^-1 1.323 · 10^-3 1.323 · 10-^2 1.323 · 10^2

Mathematics
1 answer:
Ann [662]3 years ago
7 0

21 × (6.3 · 10^-4)

= (21 × 6.3) × 10^-4

= 132.3 × 10^-4

= 1.323 × 10^-2

Answer

1.323 × 10^-2

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Answer:

Step-by-step explanation:

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Approximately how many inches are in 2500 millimeters? Given 1 in = 2.54 cm, 1 ft = .3048 m, 1 mi = 1.61 km.
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Let's begin with 2500 mm and convert this to cm.  

2500 mm = 250 cm

Next, convert 250 cm to inches.  Recall that 1 inch = 2.54 cm.

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7 0
3 years ago
Given a rectangle with length of (2x+9)cm and width of (3x+1)cm.Two squares, each with sides x cm is removed from the rectangle.
il63 [147K]

Answer: The length is 13cm and the width is 7cm

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A = W*L

In this case we have:

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Then the area of the rectangle is:

A = (2*x + 9)*(3*x + 1) cm^2

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now we remove two squares with sides of x cm

The area of each one of these squares is (x cm)*(x cm)  = x^2 cm^2

Then the area of the figure will be:

area = (6*x^2 + 29*x + 9) cm^2 - (2*x^2 ) cm^2

area = (4*x^2 + 29*x + 9) cm^2

Now we know that the area of this shape is 83 cm^2, then we need to solve:

83 cm^2 = (4*x^2 + 29*x + 9) cm^2

0 =  (4*x^2 + 29*x + 9) cm^2 - 83 cm^2

0 = (4*x^2 + 29*x - 74) cm^2

Then we need to solve:

0 = 4*x^2 + 29*x - 74

Here we can use Bhaskara's equation, the solutions of this equation are given by:

x = \frac{-29 \pm \sqrt{29^2 - 4*4*(-74)}  }{2*4} = \frac{-29 \pm 45}{8}

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The other solution is:

x = (-29 + 45)/8 = 2

x = 2

Then the length and width of the rectangle are:

Length = (2*2 + 9)cm = 13 cm

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