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Nana76 [90]
4 years ago
13

Unfortunately, no one buys your bracelets at a price of $10. this is a signal for you to _____. make more bracelets lower your p

rice increase your costs
Mathematics
2 answers:
Dmitriy789 [7]4 years ago
7 0

Answer:

Base price of one bracelet = $10

The problem which is being faced : No one is buying the bracelet at this cost

Now, to resolve this issue or problem a decision need to be taken in order to make people buy your bracelet.

The given options are :

Make more bracelets : Making more bracelets has no connection with the people not buying your bracelet. This would be effective in case there is a shortage for bracelets.

Lower your price : This could be effective as there is a possibility that people are finding it expensive to buy a brace;let at a base price of $10. So if you lower the base price of the bracelet, The chance are there that people will start buying your bracelets.

Increase your cost : Again there is no relation between increasing the cost and the the people not buying your bracelet. So, if you increase the base price of the bracelet then it will not have any effect on sale of the bracelets.

Hence, the most effective decision or person not buying a bracelet is a signal to lower the price of the bracelet.

tamaranim1 [39]4 years ago
3 0
I would think you would lower ur price. But I wouldn't  make the price so low so that it costs the same to produce them as it does to sell them, because then there would be no profit.
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34kurt

7____ is the one that couldn't used to complete a table of eqiuvalent ratios


7 0
3 years ago
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Let y(????)y(t) be a solution of y˙=17y(1−y7)y˙=17y(1−y7) such that y(0)=14y(0)=14. Determine lim????→[infinity]y(????)limt→[inf
shtirl [24]

Answer:

Step-by-step explanation:

Given that,

y' = 17y ( 1-y^7)

Let y=1

Then, y' = 0 for all t

Then show that it is the only stable equilibrium point so that as y→1, t→∞ with any initial value.

So, the graph solution will be

y(0) = 1 and this will be an horizontal line

If, y(0) > 1 then, y' < 0 by inspecting the first equation, so the graph is has decreasing solution.

Likewise, if y(0) < 1 then, y' > 0 and the graph is increasing.

So no matter the initial condition, graph of the solution will be asymptotic to the horizontal line above.

This make the limit be 1.

This shows that x = 1 is a stable equilibrium.

6 0
3 years ago
Find the distance travelled by Rubina if she takes 4 rounds of rectangular park whose length and breadth is 45m and 30m
ANTONII [103]
  • Length (l) = 45 m
  • Breadth (b) = 30 m
  • We know, perimeter of a rectangle = 2(length + breadth)
  • Therefore, perimeter of the rectangle
  • = 2(I + b)
  • = 2(45 + 30) m
  • = 2 × 75 m
  • = 150 m
  • So, the distance travelled by Rubina
  • = 4 × 150 m
  • = 600 m

<u>Answer:</u>

<em><u>The </u></em><em><u>distance </u></em><em><u>travelled</u></em><em><u> </u></em><em><u>by </u></em><em><u>Rubina </u></em><em><u>is </u></em><em><u>6</u></em><em><u>0</u></em><em><u>0</u></em><em><u> </u></em><em><u>m.</u></em>

Hope you could get an idea from here.

Doubt clarification - use comment section.

5 0
3 years ago
Please help with this limit, I have tried everything but this won't work. <br>​
Vera_Pavlovna [14]

You're lucky that I was able to do this :)

The answer is \huge\boxed{\frac{1}{4}}

8 0
3 years ago
Terry and Callie do word processing. For a certain prospectus Callie can prepare it two hours faster than Terry can. If they wor
Dahasolnce [82]

Time taken by jerry alone is 10.1 hours

Time taken by callie alone is 8.1 hours

<u>Solution:</u>

Given:- For a certain prospectus Callie can prepare it two hours faster than Terry can

Let the time taken by Terry be "a" hours

So, the time taken by Callie will be (a-2) hours

Hence, the efficiency of Callie and Terry per hour is \frac{1}{a-2} \text { and } \frac{1}{a} \text { respectively }

If they work together they can do the entire prospectus in five hours

\text {So, } \frac{1}{a-2}+\frac{1}{a}=\frac{1}{5}

On cross-multiplication we get,

\frac{a+(a-2)}{(a-2) \times a}=\frac{1}{5}

\frac{2 a-2}{(a-2) \times a}=\frac{1}{5}

On cross multiplication ,we get

\begin{array}{l}{5 \times(2 a-2)=a \times(a-2)} \\\\ {10 a-10=a^{2}-2 a} \\\\ {a^{2}-2 a-10 a+10=0} \\\\ {a^{2}-12 a+10=0}\end{array}

<em><u>using quadratic formula:-</u></em>

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

x=\frac{12 \pm \sqrt{144-40}}{2}

\begin{array}{l}{x=\frac{12 \pm \sqrt{144-40}}{2}} \\\\ {x=\frac{12 \pm \sqrt{104}}{2}} \\\\ {x=\frac{12 \pm 2 \sqrt{26}}{2}} \\\\ {x=6 \pm \sqrt{26}=6 \pm 5.1} \\\\ {x=10.1 \text { or } x=0.9}\end{array}

If we take a = 0.9, then while calculating time taken by callie = a - 2 we will end up in negative value

Let us take a = 10.1

So time taken by jerry alone = a = 10.1 hours

Time taken by callie alone = a - 2 = 10.1 - 2 = 8.1 hours

3 0
3 years ago
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