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arsen [322]
3 years ago
11

Find the diameter and circumference of a circle with the radius of 13.1 km.

Mathematics
1 answer:
Alexandra [31]3 years ago
4 0
R = 13.1
so
<span>diameter  = 13.1 x 2 = 26.2 km
</span>circumference = 2 pi r = 26.2 pi km
or
circumference = 26.2 x 3.14 = 82.27 km
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Condense the expression to the logarithm of a single quantity.
kherson [118]

\dfrac 12 \log_3 x - 2\log_3 (y +8)\\ \\=\log_3 x^{\tfrac 12}- \log_3(y+8)^2~~~~~~~~~~~~~~~~~~;[\log_b m^n = n \log_b m]\\\\=\log_3 \left( \dfrac {x^{\tfrac 12}}{y+8)^2} \right)~~~~~~~~~~~~~~~~~~~~~~~~;\left[\log_b \left( \dfrac mn \right) = \log_b m - \log_b n \right]

3 0
2 years ago
How does the graph of y = sec(x + 3) – 7 compare with the graph of y = sec(x)?
Alla [95]

Answer:

The function y = sec(x) shifted 3 units left and 7 units down .

Step-by-step explanation:

Given the function: y = sec(x)

  • If k is any positive real number, then the graph of f(x) - k is  the graph of y = f(x) shifted downward k units.
  • If p is a positive real  number, then the  graph of f(x+p) is  the graph of y=f(x)  shifted to the left  p units.

The function y = \sec(x+3)-7 comes from the base function y= sec(x).  

Since 3 is added added on the inside, this  is a horizontal shift Left 3 unit, and since 7 is subtracted on the outside, this is a vertical shift  down 7 units.

Therefore, the transformation on the given function is shifted  3 units left and 7 units down


3 0
3 years ago
Read 2 more answers
What is 5 + 11g when g = 4? A. 119, B. 64, C. 49, D. 44
SSSSS [86.1K]
The answer is C. Because if g equals 4 you multiply 11 times 4 to get 44 then it is a quite simple problem 44 + 5 = 49
8 0
2 years ago
Read 2 more answers
Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat
nirvana33 [79]

Answer:

V = \pi r^2 h

For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

\frac{dV}{dt}= \frac{100 L}{min}

Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}

And replacing we got:

\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}

And that represent 0.127 \frac{cm}{min}

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

V = \pi r^2 h

For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

\frac{dV}{dt}= \frac{100 L}{min}

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}

And solving for \frac{dh}{dt} we got:

\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}

We need to convert the rate given into m^3/min and we got:

Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}

And replacing we got:

\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}

And that represent 0.127 \frac{cm}{min}

5 0
4 years ago
Write fourteen billion in scientific notation
nikdorinn [45]

Answer:

3.4 x 107

Step-by-step explanation:

6 0
3 years ago
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