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Alinara [238K]
3 years ago
13

Please help!! thank you

Mathematics
1 answer:
uysha [10]3 years ago
6 0

Answer: It would be E

Step-by-step explanation:

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-8 / 1/2


-8 / 1/2 = 


B. -16
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Step-by-step explanation:

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Jordie builds a model from 9:14am to 9:47am how many minutes does jordie spends building his model
egoroff_w [7]

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33 minutes

Step-by-step explanation:

You would do 9:47 - 9:14am which would equal 33(minutes).

3 0
3 years ago
The rate of change in the number of bacteria in a culture is proportional to the number present. In a certain laboratory experim
mezya [45]

Answer:

(a) N(t)=10000e^{(\frac{ln5}{10})t }

(b) 25,000

(c) 4.3068 min.

Step-by-step explanation:

Rate of change in the number of bacteria is proportional to the number present.

Let N is the population of bacteria.

\frac{dN}{dt} ∝ N ⇒ \frac{dN}{dt}=kt { k = proportionality constant}

initial population No. = 10,000

                          N(t_{1} ) = 20,000

         and N(t_{1}+10 )=100,000

(a) For population growth

N(t)=N_{0}e^{kt}=10000e^{kt}

N(t_1)=10,000e^{kt_1}=20,000

e^{kt}=2

ln(e^{kt_1})=ln(2)

kt_1=ln(2)

t_{1}=\frac{ln2}{k} ----------(1)

N(t_1+t_{10})=100,000

100,000=10,000e^{k(t_1+10)}

10=e^{k(t_1+10)}

ln10=ln[e^{k(t_1+10)}]

k(t_1+10)=ln10

k(t_1)=ln10-10k

t_1=\frac{ln10-10k}{k} ----------(2)

from equation (1) and (2)

\frac{ln_2}{k}=\frac{ln10-10k}{k}

ln10-ln2=10k

k=\frac{ln5}{10}

so expression will be

N(t)=10000e^{(\frac{ln5}{10})t }

(b) for t = 20

    N_{(20)}=10,000e\frac{ln5}{10}\times 20

             =  10,000\times e^{2ln5}

            = 10,000 × 25

            = 25,000

(c) Since t_1=\frac{ln2}{k}   [from equation (1)]

                  =\frac{ln2}{\frac{ln5}{10} }

                  =\frac{ln2}{ln5}\times 10

                  = 4.3068

                 = 4.3068 min.

6 0
3 years ago
(7^5)^7 HURYYYYY................
garri49 [273]

Answer:

Exact Form:

7^{35}

Decimal Form:

3.78818692 x 10^{29}

I hope this helped! :) If this is correct please give me brainliest!

5 0
3 years ago
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