Answer:

Step-by-step explanation:
You know how subtraction is the <em>opposite of addition </em>and division is the <em>opposite of multiplication</em>? A logarithm is the <em>opposite of an exponent</em>. You know how you can rewrite the equation 3 + 2 = 5 as 5 - 3 = 2, or the equation 3 × 2 = 6 as 6 ÷ 3 = 2? This is really useful when one of those numbers on the left is unknown. 3 + _ = 8 can be rewritten as 8 - 3 = _, 4 × _ = 12 can be rewritten as 12 ÷ 4 = _. We get all our knowns on one side and our unknown by itself on the other, and the rest is computation.
We know that
; as a logarithm, the <em>exponent</em> gets moved to its own side of the equation, and we write the equation like this:
, which you read as "the logarithm base 3 of 9 is 2." You could also read it as "the power you need to raise 3 to to get 9 is 2."
One historical quirk: because we use the decimal system, it's assumed that an expression like
uses <em>base 10</em>, and you'd interpret it as "What power do I raise 10 to to get 1000?"
The expression
means "the power you need to raise 10 to to get 100 is x," or, rearranging: "10 to the x is equal to 100," which in symbols is
.
(If we wanted to, we could also solve this:
, so
)
Answer:
(3X8) - 4 + 2 - 12
Step-by-step explanation:
Following the order of operations solve:
(24) - 4 + 2 - 12
20 + 2 - 12
22 - 12
10
Don't feel bad, this is an odd question and the parenthesis do not seem to be required. Following the order of operations would result in the same answer anyway...
Odd.
Answer:
The correct options are A, B, C and D.
Step-by-step explanation:
A figure said to be congruent if:
Two figures are said to be congruent if they have same size and same shape.
If two figure are congruent that means the corresponding sides will also be congruent.
If two figure are congruent that means the the corresponding angles will also be congruent.
Now consider the provided option.
By the above definition it is clear that all the options are correct.
Hence, the correct options are A, B, C and D.
Need a picture! Then I will be happy to help :
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