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LenaWriter [7]
3 years ago
9

Expand the logarithm. log4 (3xyz​)^2

Mathematics
2 answers:
Paladinen [302]3 years ago
8 0

Answer:

Answer is (2log2 + 2log3 + 2logx + 2logy + 2logz)

Step-by-step explanation:

log4(3xyz)²

= log4 + log(3xyz)²

= log4 + 2log(3xyz)

= log(2)² + 2[log3 + logx + logy + logz]

= 2log2 + 2log3 + 2logx + 2logy + 2logz

tangare [24]3 years ago
4 0
2<span>(<span>log4</span><span>(3)</span>+<span>log4</span><span>(x)</span>+<span>log4</span><span>(y)</span>+<span>log4</span><span>(z)</span><span>)</span></span>
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Find the eqation of the line passing through the points(3,3) and (4,5)
Nikitich [7]

Answer:

The Answer is: y = 2x - 3

Step-by-step explanation:

Given points: (3, 3) and (4, 5)

Find the slope, m:

m = y - y1/(x - x1)

m = 3 - 5/(3 - 4)

m = -2/-1 = 2

Use the Point Slope form of the equation:

y - y1 = m(x - x1)

y - 5 = 2(x - 4)

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2 years ago
Four more than the quotient of a number and 3 is 9
barxatty [35]

The algebraic expression of "Four more than the quotient of a number and 3 is 9" is \frac{x}{3}+4=9

The value of x=15

Step-by-step explanation:

We need to express "Four more than the quotient of a number and 3 is 9" as algebraic expression.

Let the number = x

Solving:

quotient of a number and 3: \frac{x}{3}

Four more than the quotient of a number and 3: \frac{x}{3}+4

Four more than the quotient of a number and 3 is 9: \frac{x}{3}+4=9

Now, finding value of x:

\frac{x}{3}+4=9\\Adding\,\,-4\,\,on\,\,both\,\,sides:\\\frac{x}{3}+4-4=9-4\\\frac{x}{3}=5\\Multiply\,\,both\,\,sides\,\,by\,\,3:\\x=5*3\\x=15

So, value of x = 15

Keywords: Algebraic expression

Learn more about algebraic expression at:

  • brainly.com/question/1600376
  • brainly.com/question/1617787
  • brainly.com/question/9720317

#learnwithBrainly

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2 years ago
How many flowers spaced every 4 inches are needed to surround a circular garden with a 15-foot radius? Round all circumference a
charle [14.2K]

Answer:

283 flowers

Step-by-step explanation:

c=2pi*r

c = 1130.973 =1131

1131/4

282.75 = 283

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2 years ago
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