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insens350
6 days ago
5

Threatened monarch butterflies are monitored by measuring the area occupied by overwintering butterflies in Mexico. According to

data collected by World Wildlife Fund Mexico, monarch butterflies occupied 1.13 hectares one recent winter, and 4.01 hectares the following winter (one year later)1. If the population is growing geometrically and grows at the same rate, how much area should the overwintering butterflies occupy one year after that
Mathematics
1 answer:
Leona [12.1K]6 days ago
6 0

Answer:

14.2 hectares

Step-by-step explanation:

Fund Mexico reports that monarch butterflies occupied 1.13 hectares in one winter and 4.01 hectares the next winter (one year later)

The Geometric growth formula is

(Pt/ Po)^1/t

Where Pt = Size after t years = 4.01 hectares

Po = Initial size = 1.13

t = time = 1

=( 4.01/1.13 )^1/1

= 3.5486725664

Thus, the geometric growth rate is 3.5486725664.

The area they will occupy in hectares after one more year = Current area × Geometric growth rate

= 4.01 ×3.5486725664

= 14.230176991 hectares

Approximately equal to 14.2

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A circle with radius of \greenD{1\,\text{cm}}1cmstart color #1fab54, 1, start text, c, m, end text, end color #1fab54 sits insid
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Response:

8.86cm^2

Detailed breakdown:

Considering a rectangle measuring 3cm by 4cm with a circle of 1 cm diameter within it:

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15 days ago
Korey starts a small carwash business to save up some cash. He decides to offer two different price packages to his clients. Pac
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8$

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6 0
7 days ago
A restaurant purchased kitchen equipment on January​ 1, 2017. On January​ 1, 2019, the value of the equipment was ​$14 comma 550
babunello [11306]

Answer:

\frac{dV(t)}{dt} = - 1675.38

Step-by-step explanation:

In 2017, the kitchen equipment's value was recorded at $14,550.

V(0)=$14550

Its following value is represented by V(t)=14550e^{-0.158t.

We need to ascertain the rate of value change as of January 1, 2019.

V(t)=14550e^{-0.158t

\frac{dV(t)}{dt} =\frac{d}{dt}14550e^{-0.158t

\frac{dV(t)}{dt} =14550 \frac{d}{dt}e^{-0.158t

\\Let u= -0.158t,\frac{du}{dt}=-0.158

\frac{dV(t)}{dt} =14550 \frac{d}{du}e^u\frac{du}{dt}

\frac{dV(t)}{dt} =14550 X -0.158 e^{-0.158t}=-2298.9e^{-0.158t}

As of 2019, which is 2 years later, we set t=2.

The rate of value change is

\frac{dV(t)}{dt} =-2298.9e^{-0.158X2}

=\frac{dV(t)}{dt} =-2298.9e^{-0.316}= -1675.38

3 0
1 month ago
Which expression is equivalent to StartFraction RootIndex 7 StartRoot x squared EndRoot Over RootIndex 5 StartRoot y cubed EndRo
Inessa [12141]

Answer:

Choice A.

Detailed explanation:

The provided expression is

\dfrac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}

where, y\neq 0.

Our goal is to identify an expression that matches the one given.

The current expression can be expressed as

\dfrac{(x^2)^{\frac{1}{7}}}{(y^3)^{\frac{1}{5}}} [\because \sqrt[n]{x}=x^{\frac{1}{n}}]

\dfrac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}} [\because (a^m)^n=a^{mn}]

x^{\frac{2}{7}}y^{-\frac{3}{5}} [\because a^{-n}=\dfrac{1}{a^n}]

Thus, the right choice is A.

8 0
20 days ago
Read 2 more answers
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