5/k = 15/20 what would the k be?
1 answer:
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What is the relationship between v(t), x(t), and a(t)? I assume that v(t) is a velocity and x(t) is distance traveled in relation to v(t) and a(t) is the acceleration in relation to v(t). So . . .
x(t) = ∫v(t)dt = ∫(8t + 25)dt = 4
+ C
a(t) = d/dt(v(t)) = d/dt(8t+25) = 8
x(t) = ∫v(t)dt = ∫(8t + 25)dt = 4
a(t) = d/dt(v(t)) = d/dt(8t+25) = 8
Answer:
Step-by-step explanation:
They are not.
For the <em>g[f(x)]</em> function, you substitute ³/ₓ ₋ ₁ from the <em>f</em><em>(</em><em>x</em><em>)</em><em> </em>function in for <em>x</em><em> </em>in the <em>g</em><em>(</em><em>x</em><em>)</em><em> </em>function to get this:
Then, you bring <em>x</em><em> </em><em>-</em><em> </em><em>1</em><em> </em>to the top while changing the expression to its conjugate [same expressions with opposite symbols]:
You could also do this [attaching another negative would make that positive].
For the <em>f[g(x)]</em> function, ²/ₓ from the <em>g(x)</em> function for <em>x</em><em> </em>in the <em>f(x)</em> function to get this:
Now, if you look closely, ²/ₓ is written as 2x⁻¹, and according to the Negative Exponential Rule, you bring the denominator to the numerator while ALTERING THE INTEGER SYMBOL FROM NEGATIVE TO POSITIVE:
When this happens, <em>x</em><em> </em>leaves the <em>two</em> and gets attached to the <em>three</em><em>,</em><em> </em>and 1 gets an <em>x</em><em> </em>attached to it.
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