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3 years ago

Brian head 42 class tickets to so he sold 5/6 of the tickets how many tickets did Brian sell

Mathematics

2 answers:

Lelechka [254]3 years ago

8 0

42/6=7 so 5/6 is 5*7=35. Brian sold 35 tickets.

Hope that helps.

Marta_Voda [28]3 years ago

7 0

So, he started with 42 tickets.

He sold 5/6 of them.

42 ÷ 6 =7.

7 × 5 = 35.

So, Brian sold 35 tickets.

Glad I could help, and good luck!

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Math question

strojnjashka [21]

Answer:

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

Step-by-step explanation:

The volume ( $V$ ), in cubic centimeters, and surface area ( $A_{s}$ ), in square centimeters, formulas for the candle are described below:

$V = \pi\cdot r^{2}\cdot h$ (1)

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$ (2)

Where:

$r$ - Radius, in centimeters.

$h$ - Height, in centimeters.

By (1) we have an expression of the height in terms of the volume and the radius of the candle:

$h = \frac{V}{\pi\cdot r^{2}}$

By substitution in (2) we get the following formula:

$A_{s} = 2\pi \cdot r^{2} + 2\pi\cdot r\cdot \left(\frac{V}{\pi\cdot r^{2}} \right)$

$A_{s} = 2\pi \cdot r^{2} +\frac{2\cdot V}{r}$

Then, we derive the formulas for the First and Second Derivative Tests:

First Derivative Test

$4\pi\cdot r -\frac{2\cdot V}{r^{2}} = 0$

$4\pi\cdot r^{3} - 2\cdot V = 0$

$2\pi\cdot r^{3} = V$

$r = \sqrt[3]{\frac{V}{2\pi} }$

There is just one result, since volume is a positive variable.

Second Derivative Test

$A_{s}'' = 4\pi + \frac{4\cdot V}{r^{3}}$

If $\left(r = \sqrt[3]{\frac{V}{2\pi}}\right)$ :

$A_{s} = 4\pi + \frac{4\cdot V}{\frac{V}{2\pi} }$

$A_{s} = 12\pi$ (which means that the critical value leads to a minimum)

If we know that $V = 3217\,cm^{3}$ , then the dimensions for the minimum amount of plastic are:

$r = \sqrt[3]{\frac{V}{2\pi} }$

$r = \sqrt[3]{\frac{3217\,cm^{3}}{2\pi}}$

$r = 8\,cm$

$h = \frac{V}{\pi\cdot r^{2}}$

$h = \frac{3217\,cm^{3}}{\pi\cdot (8\,cm)^{2}}$

$h = 16\,cm$

And the amount of plastic needed to cover the outside of the candle for packaging is:

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$

$A_{s} = 2\pi\cdot (8\,cm)^{2} + 2\pi\cdot (8\,cm)\cdot (16\,cm)$

$A_{s} \approx 1206.372\,cm^{2}$

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

3 0

3 years ago

How do you do this question?

IRINA_888 [86]

Answer:

∑ (-1)ⁿ⁺³ 1 / (n^½)

∑ (-1)³ⁿ 1 / (8 + n)

Step-by-step explanation:

If ∑ an is convergent and ∑│an│is divergent, then the series is conditionally convergent.

Option A: (-1)²ⁿ is always +1. So an =│an│and both series converge (absolutely convergent).

Option B: bn = 1 / (n^⁹/₈) is a p series with p > 1, so both an and │an│converge (absolutely convergent).

Option C: an = 1 / n³ isn't an alternating series. So an =│an│and both series converge (p series with p > 1). This is absolutely convergent.

Option D: bn = 1 / (n^½) is a p series with p = ½, so this is a diverging series. Since lim(n→∞) bn = 0, and bn is decreasing, then an converges. So this is conditionally convergent.

Option E: (-1)³ⁿ = (-1)²ⁿ (-1)ⁿ = (-1)ⁿ, so this is an alternating series. bn = 1 / (8 + n), which diverges. Since lim(n→∞) bn = 0, and bn is decreasing, then an converges. So this is conditionally convergent.

5 0

3 years ago

2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):

zlopas [31]

<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

$A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)$

First pair of opposite sides:

<u>Slope:</u>

$\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}$

Second pair of opposite sides:

<u>Slope:</u>

$\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}$

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

$d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\$

So the diagonals measure the same, therefore this is a rectangle.

5 0

3 years ago

1. Find 3 equivalent fraction for the following: a) 5/6 b) 7/11 c) 9/7

SOVA2 [1]

Answer:

Step-by-step explanation:

To find equivalent fractions, multiply the numerator and the denominator by the same table.

a) 5/6

$\frac{5*2}{6*2}=\frac{10}{12}\\\\\frac{5*3}{6*3}=\frac{15}{18}\\\\\frac{5*4}{6*4}=\frac{20}{24}$

b) 7/11

$\frac{7*3}{11*3}=\frac{21}{33}\\\\\frac{7*4}{11*4}=\frac{28}{44}\\\\\frac{7*5}{11*5}=\frac{35}{55}$

c)9/7

$\frac{9*2}{7*2}=\frac{18}{14}\\\\\frac{9*5}{7*5}=\frac{45}{35}\\\\\frac{9*10}{7*10}=\frac{90}{70}$

8 0

3 years ago

Si 14 lápices cuestan 30

katrin2010 [14]

Then one pencil costs about 2.14 cents.
<span>Entonces un lápiz cuesta alrededor de 2.14 centavos.</span>

8 0

3 years ago