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3 years ago

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The product of two consecutive positive odd numbers is 195.By constructing a quadratic equation and solving it, find the two num

bers
Please help

1 answer:

Tomtit [17]3 years ago

5 0

Step-by-step explanation:

x = one odd number

the second odd number is (x+2), as both are consecutive.

so,

x × (x + 2) = 195

x² + 2x = 195

x² + 2x - 195 = 0

the formula to solve this is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = 2

c = -195

x = (-2 ± sqrt(4 - 4×1×-195))/(2×1) =

= (-2 ± sqrt(4 + 4×195))/2 = (-2 ± sqrt(784))/2 =

= (-2 ± 28)/2 = -1 ± 14

x1 = -1 + 14 = 13

x2 = -1 - 14 = -15 which is invalid as the numbers are positive.

so, the two numbers are 13 and 15.

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Arrange the cones in order from lease volume to greatest volume

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Answer:

Volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let $V_{2}. V_{3}. and\ V_{4}.$ be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

$d_{1} = 20\ and\ h_{1}=12$

$d_{2} = 18\ and\ h_{2}=10$

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

$V=\pi r^{2} \frac{h}{3}$ ----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for $d_{1} = 20\ and\ h_{1}=12$

$r_{1} = \frac{d_{1}}{2}$

$r_{1} = \frac{20}{2}=10\ units$

$V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}$

$V_{1}=\pi (10)^{2} \frac{12}{3}$

$V_{1}=\pi\times 100\times 4$

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Similarly, for volume of the cone for $d_{2} = 18\ and\ h_{2}=10$

$r_{2} = \frac{d_{2}}{2}$

$r_{2} = \frac{18}{2}=9\ units$

$V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}$

$V_{2}=\pi (9)^{2} \frac{10}{3}$

$V_{2}=\pi\times 81\times \frac{10}{3}$

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Similarly, for volume of the cone for $r_{3} = 10\ and\ h_{3}=9$

$V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}$

$V_{3}=\pi (10)^{2} \frac{9}{3}$

$V_{3}=\pi\times 100\times 3$

$V_{3}=\pi\times 300$

$V_{3}=300\pi\ units^{3}$

Similarly, for volume of the cone for $r_{4} = 11\ and\ h_{4}=9$

$V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}$

$V_{4}=\pi (11)^{2} \frac{9}{3}$

$V_{4}=\pi\times 121\times 3$

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$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

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Answer:

please see explanation.

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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