1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ede4ka [16]
3 years ago
6

Telephone come

Mathematics
1 answer:
lbvjy [14]3 years ago
8 0

Answer:                                             zgfbnzbzdbzdfbgzbnbgfgfgfbv bzgfbvb

Step-by-step explanation:

You might be interested in
Y = -6(x – 3)2 – 9<br> What is the vertex? *
Murljashka [212]
It’s not a comic section
4 0
4 years ago
Read 2 more answers
3. A rare species of aquatic insect was discovered in the Amazon rainforest. To protect the species, environmentalists declared
navik [9.2K]

The number of months until the insect population reaches 40 thousand is 14.29 months and the limiting factor on the insect population as time progresses is 250 thousands.

Given that population P(t) (in thousands) of insects in t months after being transplanted is P(t)=(50(1+0.05t))/(2+0.01t).

(a) Firstly, we will find the number of months until the insect population reaches 40 thousand by equating the given population expression with 40, we get

P(t)=40

(50(1+0.05t))/(2+0.01t)=40

Cross multiply both sides, we get

50(1+0.05t)=40(2+0.01t)

Apply the distributive property a(b+c)=ab+ac, we get

50+2.5t=80+0.4t

Subtract 0.4t and 50 from both sides, we get

50+2.5t-0.4t-50=80+0.4t-0.4t-50

2.1t=30

Divide both sides with 2.1, we get

t=14.29 months

(b) Now, we will find the limiting factor on the insect population as time progresses by taking limit on both sides with t→∞, we get

\begin{aligned}\lim_{t \rightarrow \infty}P(t)&=\lim_{t \rightarrow \infty}\frac{50(1+0.05t)}{2+0.01t}\\ &=\lim_{t \rightarrow \infty}\frac{50(\frac{1}{t}+0.05)}{\frac{2}{t}+0.01}\\ &=50\times \frac{0.05}{0.01}\\ &=250\end

(c) Further, we will sketch the graph of the function using the window 0≤t≤700 and 0≤p(t)≤700 as shown in the figure.

Hence, when the population P(t) (in thousands) of insects in t months after being transplanted by P(t)=(50(1+0.05t))/(2+0.01t) then the number of months until the insect population reaches 40 thousand 14.29 months and the limiting factor on the insect population is 250 thousand and the graph is shown in the figure.

Learn more about limiting factor from here brainly.com/question/18415071.

#SPJ1

8 0
2 years ago
Scott has jaunt baked 24 cookies he want to put them in a paper bags in groups of 5 how many groups of 5 can he make how many co
kondaur [170]

Answer:

He can make 4 groups as for the 5th group he left with 4 cookies.

So your answer would be 4 groups and 4 cookies.

hope it helps!

6 0
3 years ago
Precise approximation of squar root of 12
Lady bird [3.3K]

3.46410161514. i used a calculator so hopefully this is correct

6 0
4 years ago
What is the perimeter of polygon LMNPQ?
Nataliya [291]

Answer:7+2/

Step-by-step explanation:

4 0
3 years ago
Other questions:
  • How to write multiplication stories?
    14·2 answers
  • A part selected for testing is equally likely to have been produced on any one of six cutting tools.
    6·1 answer
  • What is the slope of the line x = 4y - 24? Show your work. [Please help me here I'm dying I'm giving 65 points]
    11·2 answers
  • Carson and Jayden kept track of how many miles they ran during one week. Carson ran a total of 43 miles, which was 12 miles more
    7·2 answers
  • Im honestly too lazy to even attempt to do math anymore. can anyone help me with this question?​
    14·1 answer
  • E
    9·1 answer
  • Pls help geometry is hard
    12·1 answer
  • ??????????????????????
    12·1 answer
  • 100 students had some homework
    5·2 answers
  • Solve 30% of blank =90
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!