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Ksju [112]
2 years ago
9

Manuel has 18 yards of fabric to make table runners. It takes Three-fourths of a yard to make each runner. The expression that r

epresents the amount of fabric left after making t table runners is 18 minus three-fourths t. Which are possible numbers of table runners that Manuel could make? Select three options. 20 22 24 26 28.
Mathematics
1 answer:
stira [4]2 years ago
3 0

Inequality helps in comparison. The possible number of table runners that Manuel could make is 20, 22, and 24.

<h3>What is Inequality?</h3>

Inequality helps us to compare two unequal expressions to form an equation.

Given to us

Left over cloth = 18 - (3/4)t

In the given equation, t shows the number of table runners that can be made with 18 yards of fabric.

As the total amount of fabric that is used for making must not be more than 18 yards, we can write the equation as,

\frac{3}{4}t \leq  18

solving it for t,

t \leq  18 \times \dfrac{4}{3}\\\\t \leq 24

Thus, the number of tables runner that can be made with 18 yards of cloth such that the size of the table runner is \frac{3}{4} t must be equal to less than 24.

Hence, the possible number of table runners that Manuel could make is 20, 22, and 24.

Learn more about Inequality:

brainly.com/question/19491153

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2. 3 times the sum of a number x and 13
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Answer:

(x+13)2.3                 if simplifying.    

Step-by-step explanation:

sum means add. so x plus 13  times 2.3     or    2.3(x+13)

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3 years ago
Particle P moves along the y-axis so that its position at time t is given by y(t)=4t−23 for all times t. A second particle, part
sergey [27]

a) The limit of the position of particle Q when time approaches 2 is -\pi.

b) The velocity of particle Q is v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}} for all t \ne 2.

c) The rate of change of the distance between particle P and particle Q at time t = \frac{1}{2} is \frac{4\sqrt{82}}{9}.

<h3>How to apply limits and derivatives to the study of particle motion</h3>

a) To determine the limit for t = 2, we need to apply the following two <em>algebraic</em> substitutions:

u = \pi t (1)

k = 2\pi - u (2)

Then, the limit is written as follows:

x(t) =  \lim_{t \to 2} \frac{\sin \pi t}{2-t}

x(t) =  \lim_{t \to 2} \frac{\pi\cdot \sin \pi t}{2\pi - \pi t}

x(u) =  \lim_{u \to 2\pi} \frac{\pi\cdot \sin u}{2\pi - u}

x(k) =  \lim_{k \to 0} \frac{\pi\cdot \sin (2\pi-k)}{k}

x(k) =  -\pi\cdot  \lim_{k \to 0} \frac{\sin k}{k}

x(k) = -\pi

The limit of the position of particle Q when time approaches 2 is -\pi. \blacksquare

b) The function velocity of particle Q is determined by the <em>derivative</em> formula for the division between two functions, that is:

v_{Q}(t) = \frac{f'(t)\cdot g(t)-f(t)\cdot g'(t)}{g(t)^{2}} (3)

Where:

  • f(t) - Function numerator.
  • g(t) - Function denominator.
  • f'(t) - First derivative of the function numerator.
  • g'(x) - First derivative of the function denominator.

If we know that f(t) = \sin \pi t, g(t) = 2 - t, f'(t) = \pi \cdot \cos \pi t and g'(x) = -1, then the function velocity of the particle is:

v_{Q}(t) = \frac{\pi \cdot \cos \pi t \cdot (2-t)-\sin \pi t}{(2-t)^{2}}

v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}

The velocity of particle Q is v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}} for all t \ne 2. \blacksquare

c) The vector <em>rate of change</em> of the distance between particle P and particle Q (\dot r_{Q/P} (t)) is equal to the <em>vectorial</em> difference between respective vectors <em>velocity</em>:

\dot r_{Q/P}(t) = \vec v_{Q}(t) - \vec v_{P}(t) (4)

Where \vec v_{P}(t) is the vector <em>velocity</em> of particle P.

If we know that \vec v_{P}(t) = (0, 4), \vec v_{Q}(t) = \left(\frac{2\pi\cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, 0 \right) and t = \frac{1}{2}, then the vector rate of change of the distance between the two particles:

\dot r_{P/Q}(t) = \left(\frac{2\pi \cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, -4 \right)

\dot r_{Q/P}\left(\frac{1}{2} \right) = \left(\frac{2\pi\cdot \cos \frac{\pi}{2}-\frac{\pi}{2}\cdot \cos \frac{\pi}{2} +\sin \frac{\pi}{2}}{\frac{3}{2} ^{2}}, -4 \right)

\dot r_{Q/P} \left(\frac{1}{2} \right) = \left(\frac{4}{9}, -4 \right)

The magnitude of the vector <em>rate of change</em> is determined by Pythagorean theorem:

|\dot r_{Q/P}| = \sqrt{\left(\frac{4}{9} \right)^{2}+(-4)^{2}}

|\dot r_{Q/P}| = \frac{4\sqrt{82}}{9}

The rate of change of the distance between particle P and particle Q at time t = \frac{1}{2} is \frac{4\sqrt{82}}{9}. \blacksquare

<h3>Remark</h3>

The statement is incomplete and poorly formatted. Correct form is shown below:

<em>Particle </em>P<em> moves along the y-axis so that its position at time </em>t<em> is given by </em>y(t) = 4\cdot t - 23<em> for all times </em>t<em>. A second particle, </em>Q<em>, moves along the x-axis so that its position at time </em>t<em> is given by </em>x(t) = \frac{\sin \pi t}{2-t}<em> for all times </em>t \ne 2<em>. </em>

<em />

<em>a)</em><em> As times approaches 2, what is the limit of the position of particle </em>Q?<em> Show the work that leads to your answer. </em>

<em />

<em>b) </em><em>Show that the velocity of particle </em>Q<em> is given by </em>v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t +\sin \pi t}{(2-t)^{2}}<em>.</em>

<em />

<em>c)</em><em> Find the rate of change of the distance between particle </em>P<em> and particle </em>Q<em> at time </em>t = \frac{1}{2}<em>. Show the work that leads to your answer.</em>

To learn more on derivatives, we kindly invite to check this verified question: brainly.com/question/2788760

3 0
2 years ago
An equilateral triangle has a perimeter of 15x3 + 33x5 feet. What is the length of each side?
NNADVOKAT [17]

Answer:

70

Step-by-step explanation:

So I assumed that the x was multiplication...so...15x3=45 and 33x5=165...so the perimeter is 210, so you divide that by 3 and the answer is 70!

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3 years ago
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The length of a rectangle is 8 cm greater than its width. Find the dimensions of the rectangle if its area is 105 square centime
Nikitich [7]

Answer:

105 = (w+8) *w

w = 7 cm

l = 15 cm

Step-by-step explanation:

Area of a rectangle is

A = l*w

l = w+8

105 = (w+8) *w

Distribute

105 = w^2 +8w

Subtract 105 from each side

105-105 = w^2 +8w -105

0 =w^2 +8w -105

Factor

What 2 number multiply to 105 and add to 8

15*-7 = -105

15-7 = 8

(w+15) (w-7) =0

Using the zero product property

w+15 =0               w-7 =0

w=-15                      w=7

impossible

not negative

w =7

l = 15

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3 years ago
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