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nika2105 [10]
3 years ago
13

If 11x - 7= 2, then x=

Mathematics
1 answer:
Alexxx [7]3 years ago
6 0
X = 0.818181818181818.
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Please help and thank you. ​
Marina CMI [18]

Answer:

4.

  • 400 ft elevation at 1 km and 3 km from the start
  • 300 ft/km average rate of change over the first 4 km

5.

  • length = 3.5 cm
  • width = 3.5 cm
  • height = 7.0 cm

6.

  • 6 mo: $1018.14
  • 5 yrs: $1196.89
  • avg incr: $3.28 per month

Step-by-step explanation:

<h3>4.</h3>

A graphing calculator is handy for solving problems involving cubic polynomials. You're interested in where the elevation is 400 ft. Since the value of f(x) is in hundreds of feet, you want to find x such that f(x) = 4.

Values of x where that is the case are x=1 and x=3, representing distances of 1 km and 3 km from the start of the road.

The average rate of change of elevation is the difference in elevation divided by the difference in distance from the start:

  average rate of change = (f(4) -f(0) hundred ft)/((4 - 0) km)

  = (19 -7)/4 hundred feet/km

  = 12/4  hundred feet/km = 300 ft/km

___

<h3>5.</h3>

You want to find x when ...

  A(x) = 122.5 cm²

  10x² = 122.5 cm² . . . . substitute the given expression for A(x)

  x² = 12.25 cm² . . . . . . divide by 10

  x = √(12.25 cm²) = 3.5 cm . . . . take the square root

The diagram tells you ...

  length = width = x = 3.5 cm

  height = 2x = 7.0 cm

___

<h3>6.</h3>

Evaluate the given expression for the different values of m:

  $1000·1.003^6 ≈ $1018.14 . . . . 6-month value

  $1000·1.003^60 ≈ $1196.89 . . . . 5-year value

The increase is $1196.89 -1000.00 = $196.89. That increase took place over 60 months, so the average increase per month is ...

  $196.89/(60 mo) ≈ $3.28 per mo . . . . average per month over 5 years

8 0
3 years ago
Please help!!
sveta [45]
For this case we have the following equation:
 d =  \sqrt{\frac{3h}{2}}
 Where,
 d: the distance they can see in thousands
 h: their eye-level height in feet
 For Kaylib:
 d = \sqrt{\frac{3(48)}{2}}
 d=\sqrt{3(24)}
 d=\sqrt{72}
 d=6\sqrt{2}
 For Addison:
 d = \sqrt{\frac{3(85\frac{1}{3})}{2}}
 d = \sqrt{\frac{256}{2}}
 d=\sqrt{128}
 d=8\sqrt{2}
 Subtracting both distances we have:
 8\sqrt{2} - 6\sqrt{2} = 2\sqrt{2}
 Answer:
 
2\sqrt{2}
 B. 2√2 mi
6 0
3 years ago
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