That's going to depend a lot on what you want to "do" to it.
I'm going to assume that you want to find a number for 'x' that makes
the whole equation a true statement. Here's how I would do it:
<u> 5 ln(x) = 35</u>
Divide each side by 5: ln(x) = 35/5
ln(x) = 7
Raise 'e' to the power of each side: e^[ ln(x) ] = e⁷
But e^[ ln(x) ] is just 'x'. So x = e⁷ = <em>1,096.633...</em> (rounded)
That's the only number you can write in place of 'x'
and make the original equation true.
Answer:
-x^2+2x-3
Step-by-step explanation:
(x^2 - 3x) + (-2x^2 + 5x - 3)
Combine like terms
-x^2+2x-3
Answer:
write the whole number and mixed number as improper fractions i believe
Answer:
(4×5,4×5)=(20,20)
Step-by-step explanation:
formula =(
(X×a,X×b)
Answer:
top right corner
Step-by-step explanation: