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prisoha [69]
1 year ago
5

A flower vendor sells roses for 50 cents each. How much does she pay per flower is she makes $6.00 on every twenty dollars worth

sold?
Mathematics
1 answer:
Lelu [443]1 year ago
3 0

One rose is 50 cents, so 2 roses cost $1 ( 50 cents x 2).

2 roses per dollar x 20 dollars = 40 total roses sold.


When they sell $20 dollars they make $6, so that means they pay 20-6 = $14 dollars for the 40 roses.


$14 / 40 roses = 0.35 per rose.

She pays 35 cents per rose.

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A grocery bag contains x apples, each weighing of a pound, and y pounds of grapes. The total weight of the grocery bag is less t
kupik [55]
First let's write out the inequality before choosing a graph.

x apples each weighing 1/3 of a pound: 1/3x

y pounds of grapes: y

So...

1/3x + y < 5

The maximum weight is 4 pounds since the total weight of both the grapes and apples are less than 5.

In the y-axis, the first, third, and fourth graphs already exceed the capacity of 5 pounds.

So, by process of elimination, the correct graph for this problem is the second one.
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For a quadrilateral to be a parallelogram, its opposite sides and _____ must be congruent.
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<span>opposite sides and _____ </span>and angles 
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SOVA2 [1]

y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}

The arc length of the curve is

\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx

which has a value of about 5.99086.

Let f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}. Split up the interval of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [9/2, 5]

The left and right endpoints are given respectively by the sequences,

\ell_i=\dfrac{i-1}2

r_i=\dfrac i2

with 1\le i\le10.

These subintervals have midpoints given by

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4

Over each subinterval, we approximate f(x) with the quadratic polynomial

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

so that the integral we want to find can be estimated as

\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It turns out that

\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{f(\ell_i)+4f(m_i)+f(r_i)}6

so that the arc length is approximately

\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086

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Answer:

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