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Anettt [7]
3 years ago
7

A whale is 85m below the surface of the sea. A bird is directly above the whale and 28m above the surface of the sea. Find the h

eight of the bird above the whale.
Mathematics
1 answer:
Alchen [17]3 years ago
5 0
113m, since 85m + 28 = 113m
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A park is rectangular with a length of ⅔ miles. If the area of the park is 3/9 square miles, what is the width?
Zielflug [23.3K]

Answer:

Area of the rectangular park = 2/3 square mile

Let us assume the width of the rectangular park = x

Then

The length of the rectangular park = 2 2/3 * x mile

                                                      = 8x/3 mile

Then

Area of the rectangular park = Length * Width

2/3 = (8x/3) * x

2/3 = 8x^2/3

(2 * 3)/3 = 8x^2

2 = 8x^2

x^2 = 2/8

x^2 = 1/4

x^2 = (1/2)^2

x = 1/2

So the width of the park is 1/2 feet.

Step-by-step explanation:

Brainliest would be nice

4 0
3 years ago
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Expand and simplfy (2x+1)(x+3)
Nina [5.8K]

Answer:

(2x+1)(x+3)

2x(x+3)+1(x+3)

2x2+6x+x+3

2x2+7x+3 (2x2 means two x square)

if you want you can also multiply 2x+3 by x+3

like this:

(2x+1)(x+3)

x(2x+1)+3(2x+1)

2x2+x+6x+3

2x2+7x+3 (2x2 means two x square)

They will both give you the same answer.

5 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
How far did Eddie Travel in all??? 30 pts
dedylja [7]

The last dot to the right would be the total distance.

So at 6 pm he had traveled 45 km.

The answer is 45.

6 0
3 years ago
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ABC Bookstore sells new books according to the demand function P = -Q + 25, where P is the price and Q is the quantity. If the b
Ghella [55]
P = -Q + 2518 = -Q + 2518-25 = -Q-7 = -Q-7/-1 = -Q/-1Q = 7
Therefore, the store can expect to sell 7 new books if they will charge $18 for the new books. 
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3 years ago
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