Answer:
I would say getting all the them right.
Step-by-step explanation:
There isn't really a way to get more points unless you're getting the questions all correct- So I would say just grind those points!
Answer:
p(t) = 0 for t = 1
p(t) = 1 for t = 1/8 = 8^-1
Step-by-step explanation:
the graph you will have to do yourself.
just go there and type in

well, don't type "log" in letters.
you start by typing the "-" sign, and then you need to look up the functions by clicking on the "funcs" button and look for the log functions .
pick the

option. and then simply enter 8 as the first parameter in the {} brackets and x as the second in the () brackets.
and then you see.
any logarithm is 0 for x (or t) = 1.
because any a⁰ = 1.
and the logarithm gives you that exponent of the base number that leads to the given x value.
in other words : a logarithm is the inverse function of an exponential function.
the exponential function is
y = a^x
and the logarithm then determines

that is all.
and

means that the logarithm itself delivered -1.
and 8^-1 = 1/8
so, p(t) = 1 for t = 1/8
Answer:
<em>Area</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>garden</em><em>=</em><em>1</em><em>2</em><em>1</em><em> </em><em>square</em><em> </em><em>meter</em>
<em>perimeter</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>garden</em><em>=</em><em>4</em><em>4</em><em> </em><em>meter</em>
Step-by-step explanation:
Area of a square=side*side
=11*11
=<u>1</u><u>2</u><u>1</u><em><u>s</u></em><em><u>q</u></em><em><u>u</u></em><em><u>a</u></em><em><u>r</u></em><em><u>e</u></em><em><u> </u></em><em><u>meter</u></em>
Perimeter of a square garden=side*4
=11*4
=44 m
Step-by-step explanation:
-3y=12+3x
-y=4+x
y=-4-x and m which is the gradient is the coefficient of x which is -1, therefore m=-1
Here is the solution:
0.59 * 10 = $5.90
0.99 * 5 = $4.95
$5.90 - $4.95 = $0.95
You would save 0.95$ if you bought
10 candy bars at the sale price
rather than at the regular price. I am hoping that this answer has
satisfied your query and it will be able to help you in your endeavor, and if
you would like, feel free to ask another question.