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mart [117]
3 years ago
6

How do I set up 37 and 43

Mathematics
1 answer:
jarptica [38.1K]3 years ago
8 0
37)

keep in mind that the perimeter of a rectangle is length + length + width + width or P = 2l + 2w, or P = 2(l+w).

we know the perimeter of the box's width and length is 36, therefore then

\bf \stackrel{P}{36}=2(\stackrel{length}{l}+\stackrel{width}{w})\implies 18=l+w\implies \boxed{18-w=\stackrel{length}{l}}
\\\\\\
V(w)=4(w)(18-w)\implies V(w)=-4w^2+72w

check the first picture below.

now, that parabolic graph, goes up up up reaches a U-turn and the back down, so it has a "maximum" point, and that is when the volume is the highest, namely V(w).

\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
V(w) = &{{ -4}}w^2&{{ +72}}w&{{ +0}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
{{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\implies \stackrel{maximum~volume}{0-\cfrac{72^2}{4(-4)}}


43)

is pretty much the same thing, checking the vertex coordinates of the parabola, check the second picture below,

\bf h=64t-16t^2\implies h=-16t^2+64t+0\\\\\\
\textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
h = &{{ -16}}t^2&{{ +64}}t&{{ +0}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
\stackrel{\textit{it takes this many seconds}}{-\cfrac{64}{2(-16)}}\qquad \qquad \stackrel{\textit{it went up this many feet}}{0-\cfrac{64^2}{4(-16)}}

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If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
2 years ago
Help me please thanks
SVEN [57.7K]

Answer:

5/6

Step-by-step explanation:

(-6 - 4) / (3 * -4)  [substitute a and b]

-10 / -12  [multiply the values in parenthesis]

5 / 6  [simplify by dividing by the common factor 2]

6 0
2 years ago
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Can someone get the answer and explain i will give brainliest.
uranmaximum [27]
Answer: (a^2bc)^2 for the first one

6 0
3 years ago
Please show your work and check!!
ad-work [718]

Step 1. Add 6 to both sides

a ≥ 1 + 6

Step 2. Simplify 1 + 6 to 7

a ≥ 7

4 0
2 years ago
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What is the area, in square yards, of the polygon shown?
Alexeev081 [22]

Answer:46


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Label the vertices (points) of the polygon starting with 1 at an arbitrary vertex and continuing clockwise around the polygon. There should be as many vertices as there are sides. E.g. for a pentagon (five sides) there will be five vertices.


Draw a line from vertex 1 to vertex 3. This will make one triangle, with vertices 1, 2, and 3. If there are only 4 sides, it will also make a triangle with vertices 1, 3 and 4.If the polygon has more than 4 sides, draw a line from vertex 3 to vertex 5. Continue in this way until you run out of vertices.


Compute the area of each triangle. The formula for the area of a triangle is 1/2 * b * h, where b is the base and h is the height.


Add up the areas, and this is the area of the polygon.

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3 years ago
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