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Elan Coil [88]
3 years ago
5

4.

Mathematics
1 answer:
noname [10]3 years ago
8 0

Answer:

(a) Candy's initial sum as a terms of x is $6x

(b) x = $60

(c) $350

Step-by-step explanation:

The given parameters are;

The ratio in which Aliyah, Brenda and Candy share the sum of money = 3:5:6

The amount Candy later gives Aliyah = $100

The amount Candy later gives Brenda = $50

The new ratio of the sum of the shared money between Aliyah, Brenda and Candy = 2:3:3

(a) Whereby Aliyah has $3x at the start, we have;

Total sum of mony = Y

Amount of Aliyah's initial share = Y × 3/(3 + 5 + 6) = Y×3/14

Therefore, Y×3/14 = $3x

x = Y×3/14 ÷ 3 = Y/14

Amount of Candy's initial share = Y × 6/14

Therefore Candy's initial sum as a terms of x = $6x

(b) Given that Aliyah's and Candy's initial sum as a function of x are   $3x and $6x, therefore, in the ratio 3:5:6, Brenda's  initial sum as a function of x = $5x

Which gives;

Total amount of money = $14x

With

6x - 150, 3x + 100, and 5x + 50, the ratio =is 2:3:3

Therefore, we have;

14·x × 2/(2 + 3 + 3) = (6·x - 150)

14·x × 2/(8) = (6·x - 150)

14·x × 1/4 = (6·x - 150)

7·x/2 = (6·x - 150)

12·x - 300 = 7·x

12·x - 7·x = 300

5·x = 300

x = $60

(b) The final amount of money with Brenda = 5x + 50 = 5 × 60 + 50 = $350

The final amount of money with Brenda = $350.

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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

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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
A car enters a turnpike 22 miles north of the town. The car travels north at an average speed of 60 miles per hour. How far is t
QveST [7]

Answer:

y = 60x + 22

262 miles

Step-by-step explanation:

Given that the car enters the turnpike at 22 miles, his initial starting point, or initial value (b) is 22.  You also know the rate at which the car travels is 60 miles per hour, which is the slope (m).  Using the slope-intercept formula of a linear function: y = mx + b, you can plug in your initial value and rate:

y = 60x + 22, where y = the total miles and x = the number of hours traveled

After 4 hours:

y = 60(4) + 22 = 240 + 22 = 262 miles

8 0
3 years ago
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