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Elenna [48]
3 years ago
12

Karim builds a wooden table. 3 Superscript 4 in. A rectangle with a width of question mark inches. The rectangular top of his ta

ble has an area of 3 Superscript 7 square inches and a length of 3 Superscript 4 inches. What is the width of the table top?
Mathematics
2 answers:
Rasek [7]3 years ago
3 0

Answer:

  3^3 in = 27 in

Step-by-step explanation:

The area is given by the formula ...

  A = LW

Filling in the given values, we can find the unknown.

  3^7 in^2 = (3^4 in)×W

  (3^7 in^2)/(3^4 in) = W = 3^3 in . . . . . . . divide by the coefficient of W

The width of the tabletop is 3^3 = 27 inches.

___

The applicable rule of exponents is ...

  (a^b)/(a^c) = a^(b-c)

krok68 [10]3 years ago
3 0

Answer:

Step-by-step explanation:

just took the test answer is B

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Each year farmer records the number of pumpkins he sold for 5 weeks. The mean is 782 and the mean absolute deviation is 52. List
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When we have a mean M and a standard deviation S, the values that are within the mean absolute deviation are all the values in the interval:

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In this case, we have:

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Then the interval is:

[782 - 52, 782 + 52]

[730, 834]

Then any number between 730 and 834 are possible answers to this question, for example, we can choose:

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HELP!! Algebra help!! Will give stars thank u so much <333
Anna35 [415]

Answers:

  • Part a)  \bf{\sqrt{x^2+(x^2-3)^2}
  • Part b)  3
  • Part c)   2.24
  • Part d)  1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying d = \sqrt{x^2 + (x^2-3)^2} is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

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2 years ago
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