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kakasveta [241]
3 years ago
10

Describe the motion of a particle with position (x, y) as t varies in the given interval. (For each answer, enter an ordered pai

r of the form x, y.) x = 1 + sin(t), y = 3 + 2 cos(t), π/2 ≤ t ≤ 2π
Mathematics
1 answer:
DerKrebs [107]3 years ago
7 0

Answer:

The motion of the particle describes an ellipse.

Step-by-step explanation:

The characteristics of the motion of the particle is derived by eliminating t in the parametric expressions. Since both expressions are based on trigonometric functions, we proceed to use the following trigonometric identity:

\cos^{2} t + \sin^{2} t = 1 (1)

Where:

\cos t = \frac{y-3}{2} (2)

\sin t = x - 1 (3)

By (2) and (3) in (1):

\left(\frac{y-3}{2} \right)^{2} + (x-1)^{2} = 1

\frac{(x-1)^{2}}{1}+\frac{(y-3)^{2}}{4} = 1 (4)

The motion of the particle describes an ellipse.

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deff fn [24]

9x-14-2x=5(2x-6)\qquad\text{use distributive property}\\\\(9x-2x)-14=(5)(2x)+(5)(-6)\\\\7x-14=10x-30\qquad\text{add 14 to both sides}\\\\7x=10x-16\qquad\text{subtract 10x from both sides}\\\\-3x=-16\qquad\text{divide both sides by (-3)}\\\\\boxed{x=\dfrac{16}{3}}

8 0
3 years ago
Need pre-cal help. Will mark best answer brainliest
OlgaM077 [116]

so, let's keep in mind that

\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}

so let's make a quick table of those solutions, say A, B, C solutions with x,y,z liters of acid, with an acidity of 0.25, 0.40 and 0.60 respectively.


\bf \begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ A&x&0.25&0.25x\\ B&y&0.40&0.4y\\ C&z&0.60&0.6z\\ \cline{2-4}&\\ mixture&78&0.45&35.1 \end{array} \\\\\\ \begin{cases} x+y+z=78\\ 0.25x+0.4y+0.6z=35.1 \end{cases}


we know she's using "z" liters and those are 3 times as much as "y" liters, so z = 3y.


\bf \begin{cases} x+y+3y=78\\ x+4y=78\\[-0.5em] \hrulefill\\ 0.25x+0.4y+0.6(3y)=35.1\\ 0.25x+0.4y=1.8y=35.1\\ 0.25x+2.2y=35.1 \end{cases}\implies \begin{cases} x+4y=78\\\\ 0.25x+2.2y=35.1 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x+4y=78\implies \boxed{x}=78-4y \\\\\\ \stackrel{\textit{using substitution on the 2nd equation}}{0.25\left( \boxed{78-4y} \right)+2.2y=35.1}\implies 19.5-y+2.2y=35.1


\bf 1.2y=15.6\implies y=\cfrac{15.6}{1.2}\implies \blacktriangleright y=13 \blacktriangleleft \\\\\\ x=78-4y\implies x=78-4(13)\implies \blacktriangleright x=26 \blacktriangleleft \\\\\\ z=3y\implies z=3(13)\implies \blacktriangleright z=39 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{25\%}{26}\qquad \stackrel{40\%}{13}\qquad \stackrel{60\%}{39}~\hfill

5 0
3 years ago
This is a question from my sibling's math class.
Solnce55 [7]

Answer:

14

Step-by-step explanation:

You can solve using PEMDAS:

Parentheses

Exponents

Multiplication

Division

Addition

Subtraction

3(2+5)-5(3)+8=?

3(7)-5(3)+8=?

No exponents

3(7)-5(3)+8=?

21-15+8=?

No division

21-15+8=?

21-7=?

21-7=14

3(2+5)-5(3)+8=14

5 0
3 years ago
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irina1246 [14]
<h2>Maximum area is 25 m²</h2>

Explanation:

Let L be the length and W be the width.

Aidan has 20 ft of fence with which to build a rectangular dog run.

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We need to find what is the largest area that can be enclosed.

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For maximum area differential is zero

So we have

      dA = 0

      10 - 2 L = 0

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Area = 5 x 5 = 25 m²

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GenaCL600 [577]

The answer is D. 10 nickels

3 0
3 years ago
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