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Brums [2.3K]
2 years ago
5

When the sum of a number plus 3 is squared it is 11 more than the sum of the number plus 2 when squared.

Mathematics
1 answer:
forsale [732]2 years ago
5 0
x-number\\\\
(x+3)^2-11=(x+2)^2\\\\
x^2+2*3*x+3^2-11=x^2+2*2*x+4\\\\
x^2+6x+9-11=x^2+4x+4\ \ |-x^2\\\\6x-2=4x+4\\\\6x-4x=4+2\\\\2x=6\ \ |:2\\\\ x=3

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Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 6x3 ? 9x2 ? 108x + 2, [?3, 4]
SVEN [57.7K]
Assuming the question marks are minus signs
to find max, take derivitive and test 0's and endpoints

take derivitive
f'(x)=18x²-18x-108
it equal 0 at x=-2 and 3
if we make a sign chart to find the change of signs
the sign changes from (+) to (-) at x=-2 and from (-) to (+) at x=3
so a reletive max at x=-2 and a reletive min at x=3

test entpoints

f(-3)=83
f(-2)=134
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the min is at x=3 and max is at x=-2
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3 years ago
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MariettaO [177]
A=a+b
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inn [45]

Answer:

1. C. Yes, because a sum of cubes can be factored

2a. false

2b. false

2c. true

2d. false (based on what is written in the equation; refer to step-by-step)

Step-by-step explanation:

1.  Both 3 and 8 can be cubed, which is why x^3+8 can be factored (x+2)(x^2-2x+4)

2a. a^2-b^2 can be factored by the perfect square rule, so it should be (a-b)^2

2b. both terms are perfect squares, so you can factor, making it (a+b)(a-b)

2c. You can factor using the perfect square rule, making it (a+b)^2

2d. Most of what is in the equation is true, yet the correct solution would be (a-b)(a^2+ab+b^2)

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Your answer would be A
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