130 rounded to the nearest hundred

Mathematics

2 answers:

Jet001 [13]2 years ago

3 0

130 is closer to 100 than 200, so it is 100

konstantin123 [22]2 years ago

3 0

100 because why not

You might be interested in

What is the least number of cuts you would need to make to cut a 3 by 3 by 3 inch block of cheese into 1 by 1 by 1 inch blocks o

Semenov [28]

Answer:

6 Cut Minimum

Step-by-step explanation:

According to the Question,

You have a 3x3x3 cube.

First make two cuts parallel to base of the cube each at a distance of 1 unit . This leaves you with three cuboids of dimensions 3x3x1.
Now make two parallel cuts parallel to the height of the cube. This leaves you with nine cuboids of dimensions 3x1x1.
Now make the final two cuts each 1 unit apart parallel to width and you are left with 27 cubes of 1x1x1.

So , the final answer is 6 cuts.

5 0

2 years ago

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