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I am Lyosha [343]
2 years ago
12

Can anyone slove this with formula ( operations with rational number)​

Mathematics
2 answers:
Ipatiy [6.2K]2 years ago
6 0

Answer:

-3x / 4

Step-by-step explanation:

tankabanditka [31]2 years ago
3 0
The answer is -3/4 (-0.75 in decimal form, negative zero point seventy-five in word form). Why? Well, we can begin by converting the mixed number to an improper fraction:

8 x 1 = 8
8 + 1 = 9

Your final mixed number conversion is -9/8. Now we can simplify:

2/3 x -9/8

Since there are no common denominators currently listed, we can multiply as it is.

2 x -9 = -18
3 x 8 = 24

Your new fraction is -18/24. This fraction can be reduced to simplest form through dividing by 6.

-18 / 6 = -3
24 / 6 = 4

Your final fraction is -3/4.

Your final answer: 2/3 x (-1 1/8) = -3/4 or -0.75.

If you need help, let me know and I will gladly assist you.

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Which equation is y=(x+3)2+(x+4)2 rewritten in vertex form
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Vertex coordinates: (h, k)

Vertex form: y = a(x-h)^2 + k

y = 2(x+3) + 2(x+4)

Use distributive property:

y = 2((x+3)+(x+4))

Simplify:

y = 2(2x+7)

y = 4x+14

This is slope - intercept form, not vertex form. Vertex form is for quadratic equations - this is a linear equation.

Answer (in slope - intercept form):

y = 4x+14
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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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