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Nataly [62]
3 years ago
6

PLEASE HELP ME OUT IM FAILING MATH

Mathematics
2 answers:
Dovator [93]3 years ago
8 0

Answer: B. {x | -6 ≤ x ≤ 2}

Step-by-step explanation: Solve the system of inequalities.

Hope this helps you out! Let me know if you need anymore help with stuff like this! Good luck mate! ☺

-Leif-

MrMuchimi3 years ago
4 0

Answer:

Step-by-step explanation:

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Which division problem does the number line below best illustrate?
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Answer. Number one i think.

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What's bigger 12.5 or 3/20 ??
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12.5 is bigger. 
Because if we convert the fraction of 3/20 to normal numbers is equal to 0.15. <span>and</span> 12.5 is way bigger than 0.15 :3
Hope it helped.
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3 years ago
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write an equation of the line that is parallel to the line whose equation is 4y+9=2x and passes through the point (7,2)
Liula [17]
Ok so we’ll put the first equation into slope int form

4y=2x-9

y=1/2x -4/9

since the likes are parallel, the slope of the line we are trying to write an equation for is 1/2

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7 0
3 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

8 0
2 years ago
Jan ran for 1 hour and 15 minutes. If Jan can run a mile in 7.5 minutes, how many miles did she run? There are 60 minutes in an
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Answer:

10 miles

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7 0
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