K well uh i don't know what u meant but the 4th one i think so u
Answer:10
Step-by-step explanation:
Use the rules of logarithms and the rules of exponents.
... ln(ab) = ln(a) + ln(b)
... e^ln(a) = a
... (a^b)·(a^c) = a^(b+c)
_____
1) Use the second rule and take the antilog.
... e^ln(x) = x = e^(5.6 + ln(7.5))
... x = (e^5.6)·(e^ln(7.5)) . . . . . . use the rule of exponents
... x = 7.5·e^5.6 . . . . . . . . . . . . use the second rule of logarithms
... x ≈ 2028.2 . . . . . . . . . . . . . use your calculator (could do this after the 1st step)
2) Similar to the previous problem, except base-10 logs are involved.
... x = 10^(5.6 -log(7.5)) . . . . . take the antilog. Could evaluate now.
... = (1/7.5)·10^5.6 . . . . . . . . . . of course, 10^(-log(7.5)) = 7.5^-1 = 1/7.5
... x ≈ 53,080.96
Answer:
The hypotenuse to the nearest tenth is 8.1
Step-by-step explanation:
We can use the Pythagorean theorem to solve
a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse
Let a be the x leg and b be the y leg
a = 7 units
b= 4 units
7^2 + 4^2 = c^2
49+ 16 = c^2
65 = c^2
Take the square root of each side
sqrt(65) = sqrt(c^2)
8.062257748 =c
To the nearest tenth
8.1 =c
