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lesya692 [45]
3 years ago
15

What is the value number in the tenths place ? 1.793

Mathematics
2 answers:
Pie3 years ago
6 0

Answer:

7

Step-by-step explanation:

S_A_V [24]3 years ago
6 0

Answer:

0.7

rfdtegwyudhu

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Write 2.2 as a mixed number.
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A geometric sequence has an initial value of 9 and a common ratio of 5. What function could represent this situation?
leonid [27]

Answer:

The functions that represent this situation are A and C.

Step-by-step explanation:

A geometric sequence goes from one term to the next by always multiplying or dividing by the same value.

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This is the explicit formula for the geometric sequence whose first term is k and common ratio is r

6 0
3 years ago
HELP!! What is the approximate measure of angle x in the triangle shown?
Alex Ar [27]
<h3>Answer:   D)  130.5 degrees</h3>

=================================================

Work Shown:

c^2 = a^2 + b^2 - 2*a*b*\cos(C)\\\\10^2 = 5^2 + 6^2 - 2*5*6*\cos(x)\\\\100 = 25 + 36 - 60*\cos(x)\\\\100 = 61 - 60*\cos(x)\\\\100 - 61 = - 60*\cos(x)\\\\39 = - 60*\cos(x)\\\\\cos(x) = \frac{39}{-60}\\\\\cos(x) = -0.65\\\\x = \arccos(-0.65)\\\\x \approx 130.5416\\\\x \approx 130.5\\\\

Note: I used the law of cosines. Make sure your calculator is in degree mode.

6 0
2 years ago
I need help pleaseee!!
professor190 [17]

answer:



x=-4/7 and the slope is 7
4 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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