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3 years ago

What is equivelent to 4.05 repeating

1 answer:

8 0

$x=4.\overline{05}\\\\x=4.050505...\qquad|\cdot100\\\\100x=405.050505...\\\\100x-x=405.050505...-4.050505...\\\\99x=401\qquad|:99\\\\x=\dfrac{401}{99}$

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Simplify negative 4 and 1 over 6 − negative 6 and 1 over 3. Which of the following is correct?

Pavlova-9 [17]

The correct answer is 2 1/6.

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3 years ago

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Find f(-1) if f(x) = -2x^2 -x

kykrilka [37]

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2 years ago

Which value is in the solution set of |x|+9=-2

Eduardwww [97]

Answer:

The equation has no solutions

Step-by-step explanation:

If we re arrange the equation a little more we will obtain the following

|x|=-11

Which is a contradiction since, the |x| is always a positive number

Attached is the representation of this function y= |x|

It means, that it doesnt matter what value 'x' has, the answer is the same number but, positive

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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

2 years ago

Find the values of the variable

Leno4ka [110]

Answer:

Step-by-step explanation:

a = 180 - 100

= 80 degrees

b = 180 - 80

= 100 degrees

c = 180 - 135

= 45 degrees

8 0

3 years ago