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dimaraw [331]
3 years ago
5

Charlotte is driving from Elk Grove, CA to West Jordan, UT. She starts at 5:54 AM and continues driving for 670 minutes. When di

d she finish?
Mathematics
2 answers:
Nezavi [6.7K]3 years ago
7 0

Answer:

5:10

Step-by-step explanation:

5:54+ 670mins=finish time

  1. change the minutes to hours (divide 670 by 60)
  2. add that to the start time

5:54 + (11 hours and 16 mins)

hours

<em>6</em>, <em>7</em>, <em>8</em>, <em>9</em>,<em> 10, 11, 12, 1, 2, 3, 4</em>, 5, 6, 7, 8, 9, 10

5 pm due to the 54+6

mins

54, 55, 56, 57, 58, 59, 60, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10

10 minutes due to 54+6

andrezito [222]3 years ago
4 0

Answer:

5:56 Pm.

Step-by-step explanation:

I got the answer because i took the driving time and converted into hours and then did the math between times...

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Find the midpoint of the line segment with the given endpoints: (-1, -6) and (-6, 5).
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Step-by-step explanation:

First, we will find the midpoint for the x - coordinate first.

Midpoint (x) = \frac{-1+(-6)}{2} \\=\frac{-1-6}{2} \\=\frac{-7}{2} \\=-3.5

Now we will find the midpoint for the y - coordinate.

Midpoint (y) = \frac{-6+5}{2} \\=\frac{-1}{2} \\=-0.5

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8 0
1 year ago
What is the measurement of angle a?
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6 0
3 years ago
g A manufacturer is making cylindrical cans that hold 300 cm3. The dimensions of the can are not mandated, so to save manufactur
sdas [7]

Answer:

The dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

Step-by-step explanation:

A cylindrical can holds 300 cubic centimeters, and we want to find the dimensions that minimize the cost for materials: that is, the dimensions that minimize the surface area.

Recall that the volume for a cylinder is given by:

\displaystyle V = \pi r^2h

Substitute:

\displaystyle (300) = \pi r^2 h

Solve for <em>h: </em>

\displaystyle \frac{300}{\pi r^2} = h

Recall that the surface area of a cylinder is given by:

\displaystyle A = 2\pi r^2 + 2\pi rh

We want to minimize this equation. To do so, we can find its critical points, since extrema (minima and maxima) occur at critical points.

First, substitute for <em>h</em>.

\displaystyle \begin{aligned} A &= 2\pi r^2 + 2\pi r\left(\frac{300}{\pi r^2}\right) \\ \\ &=2\pi r^2 + \frac{600}{ r}  \end{aligned}

Find its derivative:

\displaystyle A' = 4\pi r - \frac{600}{r^2}

Solve for its zero(s):

\displaystyle \begin{aligned} (0) &= 4\pi r  - \frac{600}{r^2} \\ \\ 4\pi r - \frac{600}{r^2} &= 0 \\ \\ 4\pi r^3 - 600 &= 0 \\ \\ \pi r^3 &= 150 \\ \\ r &= \sqrt[3]{\frac{150}{\pi}} \approx 3.628\text{ cm}\end{aligned}

Hence, the radius that minimizes the surface area will be about 3.628 centimeters.

Then the height will be:

\displaystyle  \begin{aligned} h&= \frac{300}{\pi\left( \sqrt[3]{\dfrac{150}{\pi}}\right)^2}  \\ \\ &= \frac{60}{\pi \sqrt[3]{\dfrac{180}{\pi^2}}}\approx 7.25 6\text{ cm}   \end{aligned}

In conclusion, the dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

7 0
3 years ago
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