All categories

3 years ago

Please, please help!

1 answer:

4 0

$General \: line \: equation: y=mx+b \\ \Rightarrow y = \frac{2}{x} \: and \: y = 6{ x }^{2} - 7 \: won 't \: be \: lines \: when \: graphed$

You might be interested in

HELP IM IN CLASS AND I NEED IT DONE TODAY!

Nesterboy [21]

Answer:

6 0

2 years ago

3(5+3)^2 +14<br><br><br> help me pls i cannot deal w my parents again

Scilla [17]

Answer:

206

Step-by-step explanation:

I put it in a calculator

6 0

4 years ago

Read 2 more answers

What is the length of the diagonal of a square with a side that measures 11 centimeters

timofeeve [1]

A square has 4 equal sides
use pythagroeas theorem since a square has 4 right angles

a^2+b^2=c^2
a=b because the sides are equal (11)

11^2+11^2=c^2
242=c^2
square root both sides
11√2=c

the diagonal legnth is 11√2 centimiters

4 0

3 years ago

Carla wants the total weight of the bags to be four pounds. If she can only pick one more bag, which bag weight should she pick?

pychu [463]

Answer:

1/8

Step-by-step explanation:

To find which bag weight Carla needs, the first step is to find the total weight of the bags shown in the line plot.

There are three -pound bags.

There are six -pound bags.

There are four -pound bags.

Multiply the weight and number of each bag to find the total weight of each different weight bag.

Now, add the three totals.

Finally, subtract the total weight from four pounds in order to find which weight Carla needs.

So, Carla should pick one -pound bag of candy to total four pounds.

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