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Vesnalui [34]
3 years ago
10

An said that the difference between two negative numbers must be negative. Was he right? Use examples to illustrate your answer.

Mathematics
2 answers:
butalik [34]3 years ago
7 0
Yes a negative divided by a negative is a negative and a negative divide by a positive is also negative
alexandr402 [8]3 years ago
4 0

Answer:

Yes

Step-by-step explanation:

Yes, because -1-20=-21

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Below are the functions of y = |x| and y= |x| - 9 how are the functions related?
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5 0
2 years ago
Pete borrowed $400 for 1 year. He paid back a total of $440. What was the interest rate per year?
OleMash [197]

I=PRT/100

1. Make R (rate) subject

R/100= I/PT

2. Substitute and calculate

r/100= i/pt

r/100= 40/400 × 1

(<em>4</em><em>0</em><em> </em><em>i</em><em>s</em><em> </em>from 440-400.<em>T</em><em>h</em><em>e</em><em> </em><em>i</em><em>n</em><em>t</em><em>e</em><em>r</em><em>e</em><em>s</em><em>t</em>)

r/100= 0.1

r/100×100= 0.1×100

r=10% (interest rate per year)

To confirm

I=PRT

I= 400×10/100×1

I= $40 (Interest)

5 0
3 years ago
Read 2 more answers
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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