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garri49 [273]
4 years ago
10

Suma a patru numere este 600.Primul numar este cu 12 mai mare decat al doilea si cu 12 mai mic decat al treilea.Stiind ca al pat

rulea reprezinta un sfert din suma primelor trei numere,afla cele patru numere
Mathematics
1 answer:
pashok25 [27]4 years ago
8 0
Salut!
Ai ca a + b + c +d = 600;
a = b + 12 => b = a-12;
a = c - 12 => c = a + 12;
d = (a+b+c)/3 => d = a;
In final, 4a = 600 => a = 150;
b = 138;
c = 162;
d = 150;
Bafta!
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Bridge B is 165 feet shorter than Bridge A.The combined length of the two bridges 8421 feet. Find the length of each bridge
Paladinen [302]

Answer:4128 and 4293

Step-by-step explanation:

8421-165=8256

8256/2=4128

4128+165=4293

4293+4128=8421

6 0
4 years ago
If I = prt; p = 350, r = 6%, I = 42, then t = 2 A. True B. False
Liono4ka [1.6K]

t = 2 → A true

given I = PRT

to find T , divide both sides by PR

note that R = 6% = \frac{6}{100} = 0.06

T = \frac{I}{PR} = 42/( 350 × 0.06 ) = 2



4 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
Eighteen percent of apples grown in an orchard have defects. Let X = the number of apples that are randomly inspected from the o
KonstantinChe [14]

Answer:

0.219 or 21.028%

Step-by-step explanation:

0.18 are defective, so

P(x=1) = 0.18

P(x=2) = 0.18^2 = 0.0324

P(x=3) = 0.18^3 = 0.005832

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2 years ago
Use the graph off to find the value f(5).
Rus_ich [418]

Answer:

where is the graph

Step-by-step explanation:

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3 years ago
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