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Ratling [72]
3 years ago
5

john is saving money at a constant rate. Suppose he initially has $110 saved and after 2 months he has $160 saved. at what rate

is john saving
Mathematics
2 answers:
Ivan3 years ago
5 0

The rate John is saving is 25. How? You can do 160-110 (to find difference). After this, you will get 50. You can now divide this by 2 (because the problem says two months), which is 25. You can see if this is correct by adding 25 to 110 two times.


110 + 25

= 135


135+25

=160


Allisa [31]3 years ago
4 0
He saves $25 per month
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Advocard [28]

Answer:

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Step-by-step explanation:

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2 years ago
(x^2−1)(3x−4)(3x+4) what is the product
Andrei [34K]

Answer:

9x^4-25x^2+16

Step-by-step explanation:

(x^2−1)(3x−4)(3x+4)=(*)

(x^2−1)((3x)^2−4^2) =

(x^2 - 1)(9x^2-16)=

x^2*9x^2 - x^2*16 - 1*9x^2 +16=

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6 0
3 years ago
Which line is parallel to 8x+2y=12
Nat2105 [25]

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the final solution is

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4 0
3 years ago
Read 2 more answers
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
2 years ago
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