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Alchen [17]
3 years ago
8

For a carnival game, containers are arranged in a triangular display. The top row has 1 container. The second row has 2 containe

r. the third row has 3 containers. The pattern continues until the bottom row, which has 10 containers. A contestant knocks down 29 containers on the first throw. How many containers remain?
Mathematics
2 answers:
Ede4ka [16]3 years ago
5 0
For this, you will need to know the triangle numbers. To know how many are on row ten, find the cumulative frequency (running total) of every number up to ten

1 = 1
2 = 3
3 = 6
4 = 10
5 = 15
6 = 21
7 = 28
8 = 36
9 = 45
10 = 55

If there are 55 containers and 29 are knocked down, we do
55 - 29 = 26
26 containers remain
docker41 [41]3 years ago
4 0
26 are left. Its just addition then subtraction.
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Consider the system of equations.
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The solution to the system of equations x + 2y = 1  and -3x-2y = 5 is:

x  = -3,  y = 2

The given system of equations:

x  +  2y  =  1............(1)

-3x - 2y  =  5..........(2)

This can be written in matrix form as shown:

\left[\begin{array}{ccc}1&2\\-3&-2\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}1\\5\end{array}\right]

Find the determinant of \left[\begin{array}{ccc}1&2\\-3&-2\end{array}\right]

\triangle = 1(-2) - 2(-3)\\\triangle = -2+6\\\triangle  = 4

\triangle_x = \left[\begin{array}{ccc}1&2\\5&-2\end{array}\right]\\\triangle_x = 1(-2)-2(5)\\\triangle_x = -2-10\\\triangle_x =-12

\triangle_y = \left[\begin{array}{ccc}1&1\\-3&5\end{array}\right]\\\triangle_y = 1(5)-1(-3)\\\triangle_y = 5 + 3\\\triangle_y =8

x = \frac{\triangle_x}{\triangle} \\x = \frac{-12}{4} \\x = -3

y = \frac{\triangle_y}{\triangle} \\y = \frac{8}{4} \\y = 2

The solution to the system of equations x + 2y = 1  and -3x-2y = 5 is:

x  = -3,  y = 2

Learn more here: brainly.com/question/4428059

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In Exploration 5.4.2 Question 2, what conclusion can you make about the value of the derivative at
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The value of the derivative at the maximum or minimum for a continuous function must be zero.

<h3>What happens with the derivative at the maximum of minimum?</h3>

So, remember that the derivative at a given value gives the slope of a tangent line to the curve at that point.

Now, also remember that maximums or minimums are points where the behavior of the curve changes (it stops going up and starts going down or things like that).

If you draw the tangent line to these points, you will see that you end with horizontal lines. And the slope of a horizontal line is zero.

So we conclude that the value of the derivative at the maximum or minimum for a continuous function must be zero.

If you want to learn more about maximums and minimums, you can read:

brainly.com/question/24701109

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Answer:

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because a ste and leaf plot have the value for 2

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