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3 years ago

Find the new amount of 140 fluid ounces increased by 45%

Mathematics

1 answer:

wlad13 [49]3 years ago

8 0

$\rm 140+ \frac{45*14\not0}{10\not0} =140+ \frac{63\not0}{1\not0} =140+63=\boxed{\rm203~ounces}$

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Answer:

40 cm

Step-by-step explanation:

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Find a power series representation for the function

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$f(x)=\sum^{\infty}_{n=0} \frac{(-1)^{n}}{(2n+1)!}x^{2(2n+2)} + e^{2} \sum^{\infty }_{n=0} \frac{1}{n!}3^{n}x^{n}$ is the power series representation for the function f(x) = x³sin(x) + e³ˣ⁺². This can be obtained by using power series representation of each terms, sin x, eˣ and substituting in the function.

<h3>Find the power series representation for the function:</h3>

In the question the given function is,

⇒ f(x) = x³sin(x) + e³ˣ⁺²

We know that series representation of sin x and eˣ are:

$sin x = \sum^{\infty}_{n=0} \frac{(-1)^{n}}{(2n+1)!}x^{2n+1}$

$e^{x} = \sum^{\infty }_{n=0} \frac{1}{n!}x^{n}$

⇒ $e^{3x} = \sum^{\infty }_{n=0} \frac{1}{n!}x^{n}$

= $\sum^{\infty }_{n=0} \frac{1}{n!}3^{n}x^{n}$

Substituting the series representation in the function we get,

⇒ f(x) = x³sin(x) + e³ˣ⁺²

⇒ $f(x)=x^{3}\sum^{\infty}_{n=0} \frac{(-1)^{n}}{(2n+1)!}x^{2n+1} + e^{2} \sum^{\infty }_{n=0} \frac{1}{n!}3^{n}x^{n}$

$f(x)=\sum^{\infty}_{n=0} \frac{(-1)^{n}}{(2n+1)!}x^{2(2n+2)} + e^{2} \sum^{\infty }_{n=0} \frac{1}{n!}3^{n}x^{n}$

Hence $f(x)=\sum^{\infty}_{n=0} \frac{(-1)^{n}}{(2n+1)!}x^{2(2n+2)} + e^{2} \sum^{\infty }_{n=0} \frac{1}{n!}3^{n}x^{n}$ is the power series representation for the function f(x) = x³sin(x) + e³ˣ⁺².

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1 year ago

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All square roots of -196?

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You can’t take the square root of a negative unless you’re wanting to do it with an imaginary number

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3 years ago

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1. A buoy floats 19 yards from the eastern most point of a boat and 15 yards from the western most point of a second boat. The a

Black_prince [1.1K]

The laws of cosines and law of sines can be used given that two sides

and an included angle, or two angles a side are known.

Response:

1. The other angles in the triangle formed by the buoy are approximately;

<u>31.1° and 40.9°</u>

2. Distance of the helicopter from the first island is approximately;

<u>14.5 miles</u>

<h3>How is the Law of Sines and Cosines used?</h3>

Given parameters are;

Distance of the buoy from the easternmost point of a boat = 19 yards

Distance of the buoy from the westernmost point of the other boat = 15 yards

Angle formed from the buoy to the two boats = 108°

Distance between the two boats, <em>d</em>, is given by the law of cosines, as follows;

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d = √(586 - 570·cos(108°))

By the law of Sines, we have;

$\dfrac{d}{sin(108^{\circ})} = \mathbf{\dfrac{15}{sin(Angle \ formed \ from \ the \ boat \ on \ the \ West, \ \theta_1)}}$

Which gives;

$sin(\theta_1) = \mathbf{ \dfrac{15 \times sin(108^{\circ})}{\sqrt{586 - 570 \cdot cos(108^{\circ})} }}$

The o

$\theta_1 = arcsin \left( \dfrac{15 \times sin(108^{\circ})}{\sqrt{586 - 570 \cdot cos(108^{\circ})} } \right) \approx \mathbf{31.1^{\circ}}$

The other angles formed in the triangle containing the buoy are;

θ₁ ≈ <u>31.1</u>

θ₂ ≈ 180° - 108° - 31.1° ≈<u> 40.9°</u>

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Required:

The mileage (distance travelled) of the helicopter.

Solution:

Let <em>A</em> represent the island that has an angle of elevation to the helicopter

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θ = 180° - (15° + 35°) = 130°

By the Law of Sines, we have;

$\dfrac{20}{sin(130^{\circ})} = \mathbf{ \dfrac{Distance \ from \ island \ A }{sin(35^{\circ})}}$

Which gives;

$Distance \ of \ helicopter \ from \ island \ A = \mathbf{ \dfrac{20}{sin(130^{\circ})} \times sin(35^{\circ})}$

$Mileage \ from \ island \ A = \dfrac{20}{sin(130^{\circ})} \times sin(35^{\circ}) \times cos(15^{\circ}) \approx 14.5$

The mileage of the helicopter from the first island is approximately <u>14.5 miles</u>

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2 years ago

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