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dangina [55]
3 years ago
13

HELLLPPPPPP PLLEEEEEAAAASSSEEEE!!!!!!!!!!!!

Mathematics
1 answer:
seraphim [82]3 years ago
5 0
It would be Kellen's pumpkin because :
Jermaine = 42.009
Kim = 42.19
Kellen = 42.9
And after the 9 in 42.9 there is a 0, so: 42.90
And by comparing that to the other people's pumpkins you can see that Kellen's was the heaviest
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Eu preciso de ajuda pfvr:(
belka [17]
I’m not a true percent sure but I think it’s C
8 0
3 years ago
Is my prediction right? or no
USPshnik [31]

Answer:

so does this go with what you got I would think C correct me if i'm wrong

Step-by-step When a population or group of something is declining, and the amount that decreases is proportional to the size of the population, it's called exponential decay. In exponential decay, the total value decreases but the proportion that leaves remains constant over time.

8 0
3 years ago
At which x value will f(x)=3x+3 exceed g(x)=3x+10
Anvisha [2.4K]
We write an inequality:

f(x) > g(x)

3^x + 3 > 3x + 10

3^x > 3x + 7

This equation cannot be solved using trivial methods found in high-school classes, so we resort to graphical examination.  3x+7 is a linear function while 3^x is an exponential one (with limit zero as x approaches - \infty).  We see that 3^x = 3x+7 at approximately x=2.4 and x=-2.3.

Indeed, using a computer algebra system such as the ones on modern TI calculators and on many internet sites gives equality at x=2.42, -2.31.  By observing our graph, we see that f(x) > g(x) when x > 2.42 or x < -2.31.
3 0
3 years ago
Read 2 more answers
If $3,500 is invested at a rate of 6% , compounded continuously, approximately what amount of time would be needed to have a bal
Akimi4 [234]
D.0.21 years is the answer
4 0
3 years ago
The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E
nexus9112 [7]

Answer:

The projected enrollment is \lim_{t \to \infty} E(t)=10,000

Step-by-step explanation:

Consider the provided projected rate.

E'(t) = 12000(t + 9)^{\frac{-3}{2}}

Integrate the above function.

E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt

E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c

2000=-\frac{24000}{3}+c

2000=-8000+c

c=10,000

Therefore, E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

Now we need to find \lim_{t \to \infty} E(t)

\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

\lim_{t \to \infty} E(t)=10,000

Hence, the projected enrollment is \lim_{t \to \infty} E(t)=10,000

8 0
3 years ago
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