Answer:
a) 7/4 b) 5 c) 2
Step-by-step explanation:
Logrithmic Rule for a and b
Let a, M, N be positive real numbers.
a)
logaM - logaN = loga(M/N)
log9(7) - log9(4) = log9 (7/4)
b)
logaM + logaN = logaMN
log2 (x) + log2(9) = log2(45)
x9=45
(x9)/9 = 45/9
x = 5
c)
Change of base formula.
logb(x)=logd(b)/logd(x)
x log6(5) = log6(25) divide each term by log6(5)
x log6(5) / log6(5) = log6(25) / log6(5) Cancel common factor log6(5)
x = log6(25) / log6(5)
x = log6(5^2) / log6(5)
Expand log6(5^2) by moving 2 outside the logarithm.
x = 2log6(5) / log6(5) cancel the like term log6(5)
x = 2
Answer:
y=−x−1/3x−8
Step-by-step explanation:
y=8x−1/3x+1
To find the inverse function, swap x and y, and solve the resulting equation for x.
If the initial function is not one-to-one, then there will be more than one inverse.
So, swap the variables: y=8x−1/3x+1 becomes x=8y−1/3y+1.
Now, solve the equation x=8y−1/3y+1 for y.
y=−x−1/3x−8
Sorry I was in a rush and couldn't do the fractions as formulae. Hope it helped anyways.
I would say A
hope that helps!!!
Step-by-step explanation:
9/8 × (-7/3) =
9 × -7 = -63
8 × 3 = 24
-63/24 simplify
-21/8
To solve this
problem, let us analyze this step by step. The temperature for each day is as
follows:
Water temperature
on Sunday = 78 degrees F
Water temperature
on Monday = changed by -3 degrees F
Water temperature
on Tuesday = changed by 3 degrees F
We can see that
the total change of water temperature from Sunday to Tuesday is:
-3 + 3 = 0
Therefore there
is zero overall change. There the integer which represents the temperature
change is “0”.
Since the overall
change in water temperature is zero, hence the temperature on Sunday and on
Tuesday is similar.
Water temperature
on Tuesday = 78 degrees F
