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kirza4 [7]
3 years ago
13

Please give a simple summary of log functions.

Mathematics
1 answer:
Bess [88]3 years ago
3 0
Logarithmic functions are the inverses of exponential functions. The inverse of the exponential funtion y = ax is x = ay. The logarithmic function y = logax is defined to be equivalent to the exponential equation x = ay. A logarithm is really nothing more than an exponent.
You might be interested in
20 x (-124) x 5 find the value
zaharov [31]

Answer:

-2480

Step-by-step explanation:

multiplying a positive and a negative equals a negative. -20 (20x124)

multiply the numbers and -2480 is the answer

5 0
3 years ago
The length of a rectangle is equal to triple the width. What are the dimensions of the rectangle if the perimeter is 88cm.
suter [353]
88= 2l + 2w
l=3w
88=2(3w)+2w
88= 6w+2w
88=8w
w=11
l=3(11)
l=33
width= 11
length= 33
3 0
2 years ago
I need help with the bottom part I know the top parr
Inessa05 [86]
You used distributive property because when there’s parentheses in a equation you need to distribute the number outside of the parentheses
7 0
3 years ago
HELP ASAP i dont know how to do any of this
nadya68 [22]

Answer:

Step-by-step explanation:

Circumference is= to diameter * Pi

6\pi = 6*3.14=18.84\\15\pi=15*3.14=47.1

For the ones that are radius, double it, then pi

21*2\pi = 42\pi =42*3.14=131.88\\6.7*2\pi=13.4\pi =13.4*3.14=47.076

6 0
3 years ago
Find the equation of a line that goes through the points (-3,5) and (5,-11)
Rasek [7]

(\stackrel{x_1}{-3}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{-11}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-11}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{(-3)}}}\implies \cfrac{-16}{5+3}\implies \cfrac{-16}{8}\implies -2

\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{5}=\stackrel{m}{-2}(x-\stackrel{x_1}{(-3)}) \\\\\\ y-5-2(x+3)\implies y-5=-2x-6\implies y=-2x-1

3 0
2 years ago
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