All categories

2 years ago

The solution set for x^2 -x -56=0 is A.{7,8} b. {-7} c. {8} d. {-7,8} e. {7.-8}

Mathematics

1 answer:

podryga [215]2 years ago

5 0

I believe the answer is E

You might be interested in

What is the Vaild Of 62

SpyIntel [72]

Answer: It is -62

Step-by-step explanation: It is -62 because if you find the absolute value it is always negetive .

3 0

2 years ago

Kim has 8 dollars less than twice Ricky. If

olya-2409 [2.1K]

Based on the information given, it should be noted that the correct option is C. 3x - 8.

<h3>Solving equations.</h3>

From the information given, it was stated that Kim has 8 dollars less than twice Ricky. This will be:

= (2 × x) - 8.

= 2x - 8

Also, Ricky has x dollars. Therefore, the amount that they have together will be:

= 2x - 8 + x

= 3x - 8

Learn more about equations on:

brainly.com/question/13763238

4 0

2 years ago

A cylindrical barrel has a radius of 7.6ft and height of 10.8ft. Tripling which dimensions will triple the volume of the barrel?

noname [10]

If you triple the height of a barrel the dimensions will triple i belive.

3 0

3 years ago

PLEASE HELP!!! WILL GIVE BRANLIEST

Setler79 [48]

Answer:

Step-by-step explanation:

becuase you its A

6 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

2 years ago