0.433 is already rounded to the hundredths so to round it to the nearest tenths, remove the last number.
Your answer is 0.43
Yep! They are all right! Great job, could write a bit better.... sorry
Answer:
b=2
Step-by-step explanation:
b+3b - 6 =2
4b=8
b=2
Have a good day
Let
R = Ralph's age
S = Sara's age
First statement is translated as:
S = 3R
Second statement is translated as:
S + 4 = 2(R + 4)
Use the first equation to be substituted into the second one in terms of R which is the one we are actually going to solve for Ralph's age.
Since S = 3R, then
3R + 4 = 2(R + 4)
3R + 4 = 2R + 8
3R - 2R = 8 - 4
R = 4 years old
1/3 ln(<em>x</em>) + ln(2) - ln(3) = 3
Recall that
, so
ln(<em>x</em> ¹ʹ³) + ln(2) - ln(3) = 3
Condense the left side by using sum and difference properties of logarithms:


Then
ln(2/3 <em>x</em> ¹ʹ³) = 3
Take the exponential of both sides; that is, write both sides as powers of the constant <em>e</em>. (I'm using exp(<em>x</em>) = <em>e</em> ˣ so I can write it all in one line.)
exp(ln(2/3 <em>x</em> ¹ʹ³)) = exp(3)
Now exp(ln(<em>x</em>)) = <em>x </em>for all <em>x</em>, so this simplifies to
2/3 <em>x</em> ¹ʹ³ = exp(3)
Now solve for <em>x</em>. Multiply both sides by 3/2 :
3/2 × 2/3 <em>x</em> ¹ʹ³ = 3/2 exp(3)
<em>x</em> ¹ʹ³ = 3/2 exp(3)
Raise both sides to the power of 3:
(<em>x</em> ¹ʹ³)³ = (3/2 exp(3))³
<em>x</em> = 3³/2³ exp(3×3)
<em>x</em> = 27/8 exp(9)
which is the same as
<em>x</em> = 27/8 <em>e</em> ⁹