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BARSIC [14]
4 years ago
11

Differentiating Functions of Other Bases In Exercise, find the derivative of the function.

Mathematics
1 answer:
aniked [119]4 years ago
7 0

Answer:

\dfrac{dy}{dx} = \frac{2x-3}{\ln 16(x^2 - 3x)}

Step-by-step explanation:

We are given the following in the question:

y = \ln {16}(x^2 - 3x)

We have to find the derivative of the given expression.

y = \ln 16(x^2 - 3x)\\\text{Using the log propert}\\\\\log_a b = \dfrac{\log b}{\log a}\\\\dfrac{d(x^n)}{dx} = nx^{n-1}\\\\\dfrac{d(\log x)}{dx} = \dfrac{1}{x}\\\\\text{\bold{Differentiating we get}}\\\\\displaystyle\frac{dy}{dx} = \frac{d(\ln 16(x^2-3x))}{dx}\\\\= \frac{1}{16(x^2-3x)})\frac{d(x^2-3x)}{dx}\\\\=\frac{1}{\log 16(x^2-3x)}(2x - 3)\\\\= \frac{2x-3}{\ln 16(x^2 - 3x)}

\dfrac{dy}{dx} = \frac{2x-3}{\ln 16(x^2 - 3x)}

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Which of the following steps is needed to inscribe a circle in a triangle?
xxTIMURxx [149]

Construct two angle bisectors of the triangle

Answer:

Option (D) is correct

<u>Step-by-step explanation:</u>

We have to contruct two angle bisectors of the triangle.

various steps to inscribe the circle in a triangle are

  1. First draw a triangle
  2. Draw angle bisector of one angle
  3. Draw angle bisector of another angle
  4. Extent these angle bisectors, where they meet that point will be called the incenter
  5. From incenter draw prependicular to any side of the triangle
  6. The length from the point where the prependicular touches the side of triangle to the point of incenter will be radius.
  7. Use this radius and draw the incircle.
6 0
3 years ago
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How do I find the width in a frequency distribution?
Mama L [17]
I believe it is that you multiply
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3 years ago
Mr. Akika has 30 24-cent and 33-cent stamps all told. The stamps are worth $8.91. How many of each kind of stamp does he have?
riadik2000 [5.3K]

Answer:

  • 24-cent: 11
  • 33-cent: 19

Step-by-step explanation:

I find it easiest to solve these using the variable to represent the number of the most-expensive item. Let x represent the number of 33-cent stamps. Then the total value (in cents) is ...

  33x +24(30 -x) = 891

  9x = 171 . . . . . . . . . . . . . subtract 24(30), collect terms

  x = 19 . . . . . . . . . . . . divide by 9

  30-x = 11 . . . . . . . find the number of 24-cent stamps

Mr. Akika had 11 24-cent stamps and 19 33-cent stamps.

_____

<em>Additional comment</em>

You seem to have several of these to solve. They all have a similar solution.

If we let x represent the number of higher-value items, N, the total number of items, V the total value, and v1 and v2 the individual values (v2 > v1), the equation is ...

  v2(x) +v1(N -x) = V

  x(v2 -v1) = V -v1·N

  x = (V -v1·N)/(v2 -v1) . . . . generic solution

Using this formula in the problem here, we find ...

  x = (891 -24(30))/(33 -24) = 171/9 = 19 . . . as above

7 0
3 years ago
-2r+11=13 2r=2 r= I need help
Artemon [7]

Answer:

-2r = 2 - Subtract 11 from both sides

r = -1

5 0
2 years ago
PLEASE HELP ASAP!!!!
Nadusha1986 [10]
Your answer would be D.
8 0
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