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2 years ago

1. Quadratics.

1 answer:

8 0

h(24) = 21.93, vertical intercept is 5.8, (31.76,22.95) are the vertex coordinates and the distance traveled by the shot is 73.49 feet given the equation of the path of the longest shot h(x) = -0.017x² + 1.08x + 5.8. This can be obtained by understanding the concepts of graph function.

<h3>What is the value of h(24)?</h3>

Given,

h(x) = -0.017x² + 1.08x + 5.8

Put h = 24,

h(24) = -0.017(24)² + 1.08(24) + 5.8

h(24) = -9.792 + 25.92 + 5.8

h(24) = 21.93

The height of the shot put above the ground is 21.93 feet when the shot is 24 feet horizontally from the start.

<h3>What is the value of the vertical intercept?</h3>

Vertical intercept, x = 0

h(0) = -0.017(0)² + 1.08(0) + 5.8

h(0) = 5.8

The height of the shot put above the ground is 5.8 feet at the start.

<h3>What is the values of the vertex coordinates?</h3>

vertex coordinates,

(h,k) = [(-b/2a),-(b²- 4ac)/4a]

(h,k) = [(-1.08/2(-0.017)),-((1.08)²- 4(-0.017)(5.8))/4(-0.017)]

(h,k) = [(1.08/0.034),(1.5608)/0.068)]

(h,k) = (31.76,22.95)

The maximum height attained by the shot is 22.95 feet when it is horizontally 31.76 feet away from the start.

<h3>How far from the start did the shot put strike the ground?</h3>

Put h(x) = 0,

-0.017x² + 1.08x + 5.8 = 0

Use quadratic formula for solving x,

x = (-b±√b²- 4ac)/2a

Here a = -0.017, b=1.08, c=5.8

x = [-1.08±√1.08²- 4(-0.017)(5.8)]/(2×-0.017)

x = [-1.08±√1.5608]/-0.034

x = [-1.08-1.2493]/-0.034 and x = [-1.08+1.2493]/-0.034

x = 68.509 and x = - 4.98

Distance between (68.509,0) and (- 4.98,0) = √[68.509 -(- 4.98)]² + (0-0)²

= √73.49²

= 73.49 feet

Hence h(24) = 21.93, vertical intercept is 5.8, (31.76,22.95) are the vertex coordinates and the distance traveled by the shot is 73.49 feet given the equation of the path of the longest shot h(x) = -0.017x² + 1.08x + 5.8.

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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

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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,

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