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Rasek [7]
3 years ago
9

Which equation is in standard form?

Mathematics
1 answer:
evablogger [386]3 years ago
3 0
The answer is C. 2x+3y=12
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Solve for p: -9 + p &lt; 15 <br> A. p &gt; 24 B. -p &gt; 6 C. p &lt; 24 D. -p &lt; -6
Soloha48 [4]

Answer:

C.  In the given expression -9 + p < 15 , the value of p <  24.

Step-by-step explanation:

Here, the given expression is :

-9 + p < 15

Now, solving for the value of p.

If equals are added to both sides of inequality, the inequality remains unchanged.

Now, -9 + p < 15

⇒  -9 + p  + 9 < 15  + 9                                  (adding +9 on both sides)

or,  p <  24

Hence, in the given expression -9 + p < 15  the value of  p <  24.

5 0
4 years ago
Estimate the sum or difference, Use the benchmarks 0, 1/2, 1.
Oksana_A [137]
It is B might might be wing :/
5 0
4 years ago
Yoooooooooooooooo I need help
S_A_V [24]

Answer:

13/5 which is 2.6 so A.

Step-by-step explanation: Reduce the expression, if possible, by cancelling the common factors.

6 0
3 years ago
Read 2 more answers
Simplify completely 8x3+6x2-4x over -2x
Lemur [1.5K]
Simplify 
8x^3 +6x^2 + 2
4 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.

#SPJ1

8 0
2 years ago
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