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Triss [41]
3 years ago
8

6(2x-2)< 2(6x-7) Solve?

Mathematics
1 answer:
MAXImum [283]3 years ago
8 0

Answer:

3(2x-2)<(6x-7)

6x-6<6x-7

0<-1

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The anser would be 500 with the exponet of 100 solve my multiplying
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3 years ago
What is the volume of this shape?
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6 0
2 years ago
Does the table below represent a proportional<br> relationship? Show your work to prove
dolphi86 [110]

Answer:

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7 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
Use the FOIL method to find the product below. (x + 3)(x2 – 6x)
Phoenix [80]
(x + 3) • (2x - 6x)
collect like terms
(x + 3) • (-4x)
distribute -4x through the parentheses

Answer: -4x^2 - 12x

Hope this helps!
6 0
4 years ago
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