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son4ous [18]
3 years ago
6

In x3, x is called the _____.

Mathematics
2 answers:
amid [387]3 years ago
6 0

The Correct Answer is Base

sergij07 [2.7K]3 years ago
4 0

It would be called the Base

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Ter 1: Solving Simple Equation Solve x +5 = 8.​
Andru [333]

Answer:

3

Step-by-step explanation:

Step 1:

x + 5 = 8

Step 2:

x = 8 - 5

Answer:

x = 3

Hope This Helps :)

3 0
3 years ago
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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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3 years ago
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iogann1982 [59]

9514 1404 393

Answer:

  a factor of 2 is shown, but 4 was factored out

Step-by-step explanation:

The difference of squares is factored as ...

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Then the given difference of squares factors as ...

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Each of these factors has a factor of 2 that can be factored out:

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  = (2·2)(6x +5)(6x -5)

  = 4(6x +5)(6x -5)

The first factor should be 4, not 2.

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3 years ago
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I Believe the answer is C. $ 120,000
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Answer:

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x = 1.2 / (2/3)

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