please help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

2 answers:

8 0

7,825+2,267=10,092, rounded it's 10,100. C.

94,519-54,852=40,677, rounded it's 40,700. B.

7,629+5,781=13,410

6,925-4,483=2,442

78,654+27,382=106,036. C.

murzikaleks [220]3 years ago

7 0

The first one should be c the second is b and the last one is D

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Help with question 4 please

musickatia [10]

Answer:

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Step-by-step explanation:

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6 0

3 years ago

Hunter raises beetles. in a square of compost 6 ft by 6 ft, he can have 500 beetles. how many beetles can he have if his square

Rus_ich [418]

<span>If Hunter raises beetles in a square of compost with a size of 6 ft by 6 ft and having 500 beetles, and then you are to find the number of beetles with a side length that is four times longer, then you can solve this by ratio and proportion.

A = s</span>²
A = 6²
A = 36 ft²

if s = 4*the original length, then the new area is

A = 24²
A = 576 ft²

500 beetles/36 ft² = x number of beetles/576 ft²
x = 8,000 beetles

7 0

4 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.

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