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swat32
3 years ago
12

It is necessary to

Mathematics
1 answer:
Elza [17]3 years ago
8 0
19.95 x 4 = $79.80 dollars
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Please anyone help me
docker41 [41]
<h2><u>Math</u><u> </u><u>Problem</u><u> </u></h2>

<u>If</u><u> </u><u>die</u><u> </u><u>is</u><u> </u><u>rolled</u><u> </u><u>two</u><u> </u><u>times</u><u>,</u><u> </u><u>find</u><u> </u><u>the</u><u> </u><u>probably</u><u> </u><u>of</u><u> </u><u>getting</u><u> </u><u>"</u><u>1'twice</u><u>.</u>

<h2><u>Math</u><u> </u><u>Answer</u><u> </u></h2>

<u>{1}\{36}</u>

<u>Feel</u><u> </u><u>fre</u><u> </u><u>e</u><u> </u><u>too</u><u> </u><u>ask</u><u> </u><u>me</u><u> </u>

<u>Please</u><u> </u><u>understand</u><u> </u><u>it</u>

<u>#BrainliestBunch</u>

8 0
4 years ago
Please help im really struggling what is the surface area of this shape?
Semenov [28]

Answer:

  1480 in^2

Step-by-step explanation:

We can number the surfaces so we can talk about them. Starting with the very top horizontal surface, call it #1. Then the vertical surface to its right (clockwise) is #2; the lower horizontal surface you can see is #3, and the rightmost end vertical surface is #4. The bottom horizontal surface on which the figure rests is #5, and the left vertical surface you can't see is #6. Call the front vertical L-shaped surface #7, and the back vertical L-shaped surface #8.

These can all be "unfolded" into the figure shown in the attachment. (Each grid square in the attached figure is 5 inches.)

Surfaces 1–6 constitute a rectangle that is 88 inches long and 10 inches wide. The length (88 inches) is the perimeter of the L-shaped front (and back) surfaces (7 and 8). The width is the front-to-back width of the shape, marked as 10 inches at the lower right of the given figure.

The areas of the two L-shaped surfaces can be calculated several ways. One way is to recognize the L-shape as a 20 in × 24 in rectangle with a 12 in × 15 in rectangle removed from it. Another way to calculate the area is to consider it to be two trapezoids (as cut by the dashed line JV in the attached figure).

___

The discussion above and the attached figure (net) give the information we need to calculate the surface area.

The area of the central pink net is ...

  (10 in)×(88 in) = 880 in².

The area of one golden L-shape is ...

  (20 in)×(24 in) - (12 in)×(15 in) = 480 in² -180 in² = 300 in²

Then the total surface area is that of the central (pink) rectangle and two (2) (golden) L-shapes, or ...

  total area = 880 in² + 2×300 in²

  total area = 1480 in²

_____

<em>Comment on the L as trapezoids</em>

The dimensions of the edges of the L-shapes are shown in the attachment. They are computed using the information in the given figure and by subtracting heights or lengths to find the unknown dimensions.

The upper left trapezoid has a height of 9 units and bases of 12 and 20. Its area is given by the formula

  A = (1/2)(b1 +b2)h = (1/2)(12 +20)·9 = 144 . . . . in²

The lower right trapezoid has a height of 8 units and bases of 24 and 15. Its area is given using the same formula

  A = (1/2)(24 +15)·8 = 156 . . . . in²

Then the total area of one L-shape is the sum of these areas, or ...

  L-shape area = 144 in² + 156 in² = 300 in² . . . . . same as above

7 0
3 years ago
Question 8 options: Solve for the value of X. 23X = 6
Nataly [62]

Answer:

divide both sides by 23

making it to be 0.26

3 0
4 years ago
A child builds two wooden train sets. The path of one of the trains can be represented by the function y = 2cos2x, where y repre
Svet_ta [14]

Using a trigonometric equation, it is found that it will take 2.28 minutes until the two trains are first equidistant from the child.

<h3>What is the distance of each train to the child?</h3>

The distance of the first train is:

y_1 = 2\cos^{2}x

The distance of the second train is:

y_2 = 3 + \cos{x}

<h3>When are the trains equidistant to the child?</h3>

When y_1 = y_2, hence:

2\cos^{2}x = 3 + \cos{x}

The following substitution is made:

z = \cos{x}

Hence:

2z^2 = 3 + z

2z^2 - z - 3 = 0

Which is a quadratic equation with coefficients a = 2, b = -1, c = -3, hence:

\Delta = b^2 - 4ac = (-1)^2 - 4(1)(-3) = 13

z_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{1 + \sqrt{13}}{4} = 1.5

z_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{1 - \sqrt{13}}{4} = -0.65

Then, applying the trigonometric equation, considering that -1 \leq z \leq 1 due to the range of the cosine function:

z_2 = \cos{x_2}

x_2 = \arccos{z_2} = \arccos{-0.65} = 2.28

It will take 2.28 minutes until the two trains are first equidistant from the child.

You can learn more about trigonometric equations at brainly.com/question/2088730

8 0
3 years ago
The graph of y= (x + 2)(x-2)(x + 1) is shown.
In-s [12.5K]

Answer:

Point C

Step-by-step explanation:

At point C, the graph is changing from decreasing to increasing.

7 0
2 years ago
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