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tamaranim1 [39]
3 years ago
5

The length of a ribbon is 6/7 meter. Sun Yi needs pieces measuring 1/6 meter for an art project. What is the greatest number of

pieces measuring 1/6 meter that can be cut from the ribbon? How much ribbon will be left after Sun Yi cuts the ribbon?
Mathematics
1 answer:
Strike441 [17]3 years ago
6 0

Answer:

Let's find out!

Step-by-step explanation:

So we have to set up an equation

Let x= # of ribbons that can be cut

Thus:

\frac{1}{6}x =\frac{6}{7} \\x=\frac{6}{7} *\frac{6}{1} \\x=\frac{36}{7}

So, we cannot have 36/7 ribbons! It's not possible!

So we just simplify!

x=5\frac{1}{7}

There we go! She can cut 5 ribbons off!

I'll let you figure out the last one!

Hope this helps!

P.S. Stay Safe!

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Answer:

1/3

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How many solutions does the system have? \begin{cases} 5y =15x-40 \\\\ y = 3x-8 \end{cases} ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ ​ 5y=15x−40 y=3x−8 ​ C
Ksenya-84 [330]

Answer:

(C) Infinitely many solutions

Step-by-step explanation:

Converting the lines into the from ax+by+c=0

5y=15x-40\\\Rightarrow 15x-5y-40=0

y=3x-8\\\Rightarrow 3x-y-8=0

\dfrac{a_1}{a_2}=\dfrac{15}{3}=5

\dfrac{b_1}{b_2}=\dfrac{-5}{-1}=5

\dfrac{c_1}{c_2}=\dfrac{-40}{-8}=5

So, \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}.

Hence, these lines have infinitely many solutions.

3 0
3 years ago
The radius of a cone is decreasing at a constant rate of 7 inches per second, and the volume is decreasing at a rate of 948 cubi
inessss [21]

Answer:

The height of cone is decreasing at a rate of 0.085131 inch per second.        

Step-by-step explanation:

We are given the following information in the question:

The radius of a cone is decreasing at a constant rate.

\displaystyle\frac{dr}{dt} = -7\text{ inch per second}

The volume is decreasing at a constant rate.

\displaystyle\frac{dV}{dt} = -948\text{ cubic inch per second}

Instant radius = 99 inch

Instant Volume = 525 cubic inches

We have to find the rate of change of height with respect to time.

Volume of cone =

V = \displaystyle\frac{1}{3}\pi r^2 h

Instant volume =

525 = \displaystyle\frac{1}{3}\pi r^2h = \frac{1}{3}\pi (99)^2h\\\\\text{Instant heigth} = h = \frac{525\times 3}{\pi(99)^2}

Differentiating with respect to t,

\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)

Putting all the values, we get,

\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)\\\\-948 = \frac{1}{3}\pi\bigg(2(99)(-7)(\frac{525\times 3}{\pi(99)^2}) + (99)(99)\frac{dh}{dt}\bigg)\\\\\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi} = (99)^2\frac{dh}{dt}\\\\\frac{1}{(99)^2}\bigg(\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi}\bigg) = \frac{dh}{dt}\\\\\frac{dh}{dt} = -0.085131

Thus, the height of cone is decreasing at a rate of 0.085131 inch per second.

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3 years ago
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jarptica [38.1K]

Answer:

3846.5 is the answers for the question

Step-by-step explanation:

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