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prisoha [69]
3 years ago
9

How do I find the surface area of a pyramid

Mathematics
1 answer:
Fynjy0 [20]3 years ago
5 0
Find the area of all the sides and then divide each area by two. add them all up to get your answer.

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Identify the underlined place in 83.5851. Then round the number to that place. The 8 to the right is underlined.
Marina86 [1]
We have to round the number 83.5851 to the nearest hundredth.
If the next smallest place is greater than or equal to 5 we increase the value of the digit we are rounding to by one.
83.5851 ≈ 83.59
Answer: 83.59
8 0
2 years ago
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Susan determined that the expression below is equal to 7.59 . 15.91 subtract 8.32
weqwewe [10]
The answer to your question is 7.59 so shes right

3 0
3 years ago
Please help this is urgent and I am being timed
Shkiper50 [21]
Hi there!

Pauline simplified the expression correctly.

Pauline made sure she used the distributive property and combined like terms after doing so.

Hope this helps !
5 0
3 years ago
0.75 is the square root of what value?
lina2011 [118]
0.75^2 = 0.5625.....sq rt 0.5625 = 0.75
4 0
2 years ago
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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
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