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Helen [10]
2 years ago
7

What are the zeros of the equation X2 - 12x + 11 equals 0

Mathematics
2 answers:
Snowcat [4.5K]2 years ago
7 0
x^2-12x+11=0\\x^2-x-11x+11=0\\x(x-1)-11(x-1)=0\\(x-1)(x-11)=0\iff x-1=0\ or\ x-11=0\\\\\boxed{x=1\ or\ x=11}\to\fbox{a.}
Valentin [98]2 years ago
6 0
The answer is a. x=1 or x=11. Hope this helped.
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Answer:

Step-by-step explanation:

t/4+t/3=1

(3t+4t)/12=1

7t/12=1

7t=12

t=12/7

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2 years ago
A rectangle initially has dimensions 6 cm by 7 cm. All sides begin increasing in length at a rate of 1 cm/s. At what rate is the
Morgarella [4.7K]

Answer:

  49 cm²/s

Step-by-step explanation:

The problem statement tells you what to do.

Write an equation relating A, b, h:

  A = bh . . . . . . the equation for the area of a rectangle

Differentiate with respect to t:

  dA/dt = (db/dt)h + b(dh/dt) . . . . . . . product rule

To find the rate of change after 18 seconds, you need to know the dimensions b and h after 18 seconds. Since each dimension was increasing at the rate of 1 cm/s, it is 18 cm more than it was at the beginning:

At 18 seconds,

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  h = 7 cm + 18 cm = 25 cm.

Of course, db/dt = dh/dt = 1 cm/s. Then the rate of change of area is ...

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You could write a formula for the area as a function of time and differentiate that:

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3 years ago
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Answer:

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Step-by-step explanation:

According to DeMorgan's Theorem:

(W.X + Y.Z)'

(W.X)' . (Y.Z)'

(W'+X') . (Y' + Z')

Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.

For the original function:

(W . X + Y . Z)'

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= (1 + 0) = 1

For the compliment:

(W' + X') . (Y' + Z')

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=(0 + 0) . (0 + 1)

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Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.

Without the parenthesis the compliment equation looks like this:

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Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.

Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.

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Answer:

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