I needdd helppppppp plsss

Mathematics

1 answer:

Tanya [424]3 years ago

6 0

Answer:

Step-by-step explanation:

3 - 6 = -3
Plug -3 in: $-\frac{12}{-3} + (-1)^{3}$
Simplify the fraction. Two negatives equal a positive. It should now look like this: $\frac{12}{3} + (-1)^{3}$
Simplify even more. Divide 12 by 3, and change -1 to a positive by removing the plug sign: $4 - 1^{3}$
$1^{3} = 1$
Plug 1 in: 4 - 1
4 - 1 = 3

I hope this helps!

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BabaBlast [244]

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3 0

3 years ago

A quantity P is an exponential function of time I, such that P = 160 when t = 6 and P = 150 when I = 4. Use the given informatio

Klio2033 [76]

Answer:

k = 0.032
P0 = 131.836

Step-by-step explanation:

Perhaps you want to use the points (t, P) = (4, 150) and (6, 160) to find the parameters P0 and k in the equation ...

$P(t)=P_0\cdot e^{kt}$

We know from the given points that we can write the equation as ...

$P(t)=150\left(\dfrac{160}{150}\right)^{(t-4)/(6-4)}=150\left(\dfrac{16}{15}\right)^{\frac{t}{2}-2}\\\\=150\left(\dfrac{16}{15}\right)^{-2}\times\left(\left(\dfrac{16}{15}\right)^{\frac{1}{2}}\right)^t$

Comparing this to the desired form, we see that ...

$P_0=150\left(\dfrac{16}{15}\right)^{-2}\approx 131.836\\\\e^{k}=\left(\dfrac{16}{15}\right)^{1/2}\rightarrow k=\dfrac{1}{2}(\ln{16}-\ln{15})\approx 0.0322693$