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3 years ago

Find three consecutive odd numbers the product of which is 148 more than the cube of their middle number.

Mathematics

1 answer:

Anna11 [10]3 years ago

3 0

Answer:

The numbers are $-39,-37,-35$

Step-by-step explanation:

Let

x-----> the first odd number

x+2----> the middle odd number

x+4---> the third odd number

$x(x+2)(x+4)=(x+2)^{3} +148$

$(x^{2}+2x)(x+4)=(x^{2}+4x+4)(x+2)+148\\x^{3}+4x^{2}+2x^{2}+8x=x^{3}+2x^{2}+4x^{2}+8x+4x+8+148\\x^{3}+6x^{2}+8x=x^{3}+6x^{2}+12x+156\\8x=12x+156\\4x=-156\\x=-39$

The numbers are

$x=-39$

$x+2=-39+2=-37$

$x+4=-39+4=-35$

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3 years ago

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kupik [55]

Answer:

$d_{AB}\ne d_{JK}$

$d_{AB}\ne \:d_{GH}$

$d_{GH}\ne \:d_{JK}$

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

Step-by-step explanation:

Considering the graph

Given the vertices of the segment AB

A(-4, 4)

B(2, 5)

Finding the length of AB using the formula

$d_{AB}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(2-\left(-4\right)\right)^2+\left(5-4\right)^2}$

$=\sqrt{\left(2+4\right)^2+\left(5-4\right)^2}$

$=\sqrt{6^2+1}$

$=\sqrt{36+1}$

$=\sqrt{37}$

$d_{AB}\:=\sqrt{37}$

$d_{AB}=6.08$ units

Given the vertices of the segment JK

J(2, 2)

K(7, 2)

From the graph, it is clear that the length of JK = 5 units

$d_{JK}=5$ units

Given the vertices of the segment GH

G(-5, -2)

H(-2, -2)

Finding the length of GH using the formula

$d_{GH}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(-2-\left(-5\right)\right)^2+\left(-2-\left(-2\right)\right)^2}$

$=\sqrt{\left(5-2\right)^2+\left(2-2\right)^2}$

$=\sqrt{3^2+0}$

$=\sqrt{3^2}$

$\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0$

$d_{GH}\:=\:3$ units

Thus, from the calculations, it is clear that:

$d_{AB}=6.08$

$d_{JK}=5$

$d_{GH}\:=\:3$

Thus,

$d_{AB}\ne d_{JK}$

$d_{AB}\ne \:d_{GH}$

$d_{GH}\ne \:d_{JK}$

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

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Answer:

Step-by-step explanation:

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From the information given,

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