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Grace [21]
3 years ago
9

Ruben how much is 340 times 100.

Mathematics
2 answers:
salantis [7]3 years ago
7 0
The answer for 340 * 100 is 34000
serious [3.7K]3 years ago
3 0
340x100=34,000 hope this helps
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Use logarithmic differentiation to find dy/dx
liq [111]

Answer:

dy/dx  =  (x^2 - 3)^sin x [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)]

Step-by-step explanation:

y = (x^2 - 3)^sinx

ln y = ln  (x^2 - 3)^sinx

ln y = sin x * ln (x^2 - 3)

1/y * dy/dx  =   sin x * {1 / (x^2 - 3)} * 2x + ln(x^2 - 3) * cos x

1/y dy/dx =  2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)

dy/dx  =   [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)] * y

dy/dx  =  (x^2 - 3)^sin x [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)]

7 0
3 years ago
A rectangular, above-ground swimming
gregori [183]
The answer is A. 158 lbs
3 0
3 years ago
What is the area of this figure?<br><br> Enter your answer in the box.
earnstyle [38]
To find the area you do base times height. Base is the number across,and height is the number that is up and down.So in this case you will multiply 9 and 5 and 9 and 7 and get 45 and 72.Then you will add 45 and 72 together and get 117. So the area is 117. Answer 117. Hope this helps :D.
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2 years ago
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pentagon [3]

9514 1404 393

Answer:

  A = {put, the, cake, in, basket}

  B = {is, the, cake, in, basket}

  C = {no, the, cake, in, basket}

  D = { }

Step-by-step explanation:

A = (A -B) ∪ (A∩B) ∪ (A∩C}

A = {put} ∪ {the, cake, in, basket} ∪ {the, cake, in, basket}

A = {put, the, cake, in, basket}

__

B = (B -A) ∪ (B -C) ∪ (B∩A) ∪ (B∩C)

B = {is} ∪ {is} ∪ {the, cake, in, basket} ∪ {the, cake, in, basket}

B = {is, the, cake, in, basket}

__

C = (C -B) ∪ (C∩A) ∪ (C∩B)

C = {no} ∪ {the, cake, in, basket} ∪ {the, cake, in, basket}

C = {no, the, cake, in, basket}

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D = { } . . . . not defined anywhere

8 0
3 years ago
4(3x^2y^4)^3 / (2x^3y^5)^4
tatyana61 [14]

Solving the expressions  \frac{4(3x^2y^4)^3}{(2x^3y^5)^4} we get \frac{27}{4x^{6}y^{8}}

Step-by-step explanation:

We need to Solve the expression: \frac{4(3x^2y^4)^3}{(2x^3y^5)^4}

Solving:

\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}

=\frac{4(3^3x^6y^{12})}{(2^4x^{12}y^{20})}\\=\frac{4(27x^6y^{12})}{(16x^{12}y^{20})}\\=\frac{108x^6y^{12}}{16x^{12}y^{20}}

Applying exponent rule: \frac{x^a}{x^b}=x^{a-b}

=\frac{108x^{6-12}y^{12-20}}{16}\\=\frac{27x^{-6}y^{-8}}{4}

Another exponent rule says: x^{-a}=\frac{1}{x^a}=

=\frac{27}{4x^{6}y^{8}}

So, solving the expressions  \frac{4(3x^2y^4)^3}{(2x^3y^5)^4} we get \frac{27}{4x^{6}y^{8}}

Keywords: Solving Exponents  

Learn more about Solving Exponents at:

  • brainly.com/question/13174260
  • brainly.com/question/13174254
  • brainly.com/question/13174259

#learnwithBrainly

8 0
3 years ago
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