Answer:9.9
Step-by-step explanation:
Add up the scores and dividing the total by the number of scores.
Answer: jimmy needs a new job
Step-by-step explanation:
<h3>5/6 × 3/10</h3>
<em>Multiply 5/6 by 3/10 (to do this,</em><em> multiply the numerator by the numerator </em><em>and the </em><em>denominator by the denominator).</em>
<h3>5 × 3/6 × 10</h3>
<em>Perform the</em><em> multiplications </em><em>in the </em><em>previous fraction.</em>
<h3>15/60</h3>
<em>Reduce the fraction </em><em>15/60</em><em> to its</em><em> lowest expression by extracting</em><em> and canceling 15.</em>
<h3>1/4 ≈ 0.25</h3>
I assume you mean
ANSWER
An expression for P(t) is

EXPLANATION
This is a separable differential equation. Treat M and k as constants. Then we can divide both sides by M - P to get the P term with the differential dP and multiply both sides by dt to separate dt from the P terms
Integrate both sides of the equation.

Note that for the left-hand side, u-substitution gives us

hence why

Now we use
the definition of the logarithm to convert into exponential form.
The definition is

so applying it here, we get

Exponent properties can be used to address the constant C. We use

here:

If we assume that P(0) = 0, then set t = 0 and P = 0

Substituting into our original equation, we get our final answer of
-51 - 6i AND -3 + 2i - 2(24) - 8i are correct :)