What is the simplest form of 11/5 in fractions

PLZ HELP WORTH A LOT OF POINTS AND I HAVE 10 MINUTES!!!!

Lina20 [59]

Answer:

The area after 9 years will be 1,234 km^2

Step-by-step explanation:

In this question, we are tasked with calculating what the area of a certain forest that decreases at a certain percentage would be after some years.

To answer this question, we shall be using an exponential approximation.

Now, to use this exponential approximation, we shall be needing a supporting exponential mathematical equation.

This can be written as;

A = I(1-r)^t

where A is the new area we are looking for

I is the initial area which is 1700 according to the question

r is the rate of decrease which is 3.5% = 3.5/100 = 0.035

t is time which is 9 years according to the question

We plug these values and have the following;

A = 1700(1-0.035)^9

A = 1700(0.965)^9

A = 1,233.66

This is 1,234 km^2 to the nearest square kilometer

5 0

3 years ago

An island is 1 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that i

Elanso [62]

Answer:

The visitor should run approximately 14.96 mile to minimize the time it takes to reach the island

Step-by-step explanation:

From the question, we have;

The distance of the island from the shoreline = 1 mile

The distance the person is staying from the point on the shoreline = 15 mile

The rate at which the visitor runs = 6 mph

The rate at which the visitor swims = 2.5 mph

Let 'x' represent the distance the person runs, we have;

The distance to swim = $\sqrt{(15-x)^2+1^2}$

The total time, 't', is given as follows;

$t = \dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}$

The minimum value of 't' is found by differentiating with an online tool, as follows;

$\dfrac{dt}{dx} = \dfrac{d\left(\dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}\right)}{dx} = \dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} }$

At the maximum/minimum point, we have;

$\dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} } = 0$

Simplifying, with a graphing calculator, we get;

-4.72·x² + 142·x - 1,070 = 0

From which we also get x ≈ 15.04 and x ≈ 0.64956

x ≈ 15.04 mile

Therefore, given that 15.04 mi is 0.04 mi after the point, the distance he should run = 15 mi - 0.04 mi ≈ 14.96 mi

$t = \dfrac{14.96}{6} +\dfrac{\sqrt{(15-14.96)^2+1^2}}{2.5} \approx 2..89$

Therefore, the distance to run, x ≈ 14.96 mile

6 0

3 years ago

How does the volume of an oblique cylinder change if the radius is reduced to 2/9 of its original size and the height is quadrup

vesna_86 [32]

Volume of a cylinder = π r² h

Let us assume the following values:
radius = 9
height = 10

Volume = 3.14 * 9² * 10
= 3.14 * 81 *10
Volume = 2,543.40

Changes:
radius is reduced to 2/9 of its original size = 9 x 2/9 = 2
height is quadrupled = 10 x 4 = 40

Volume = π r² h
= 3.14 * 2² * 40
= 3.14 * 4 * 40
Volume = 502.40

Original volume = 2543.40 V.S. Volume after change = 502.40

The volume of an oblique cylinder decreased when its radius was decreased to 2/9 of its original size and its height is increased 4 times.

4 0

3 years ago

Will mark branliest!!

zepelin [54]

It’s A Have Ah Good Day

6 0

3 years ago

Read 2 more answers

I need help with this

eduard

Answer:are there choises?

Step-by-step explanation:

4 0

3 years ago