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Tatiana [17]
3 years ago
6

If ab = 8 and a^2+b^2=16, then what is the value of (a+b)^2 This answer is worth 100 point.

Mathematics
1 answer:
sergejj [24]3 years ago
5 0

Answer:

32

Step-by-step explanation:

(a+b)^2 = a^2 + 2ab + b^2

(a+b)^2 = a^2 + b^2 + 2ab

(a+b)^2=16+2(8)

(a+b)^2 = 16 + 16 = 32

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Answer:

the answer is below

Step-by-step explanation:

We have the total number of possible samples = 4 * 4 = 16 (As 4 choices for each value)

in addition we have to as all sample occur with equal probability, probability of each sample = 1/16, below is samling distribution of mean

x1 x2 probabilityP(x1,x2) sample mean

190 190    1/16                 190

190 100    1/16                 145

190 272    1/16                 231

190 74    1/16                 132

100 190    1/16                 145

100 100    1/16                 100

100 272    1/16                 186

100 74    1/16                        87

272 190    1/16                 231

272 100    1/16                 186

272 272    1/16                 272

272 74    1/16                 173

74 190    1/16                 132

74 100    1/16                 87

74 272    1/16                 173

74 74    1/16                 74

Summarizing above with adding duplicate values

sample mean   probability

       74              1/16

      87               1/8

    100                1/16

    132                1/8

    145                1/8

    173                1/8

   186                1/8

   190               1/16

    231               1/8

   272               1/16

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3 years ago
Cheryl is moving to a new house. Her old house is 3 km from her new house. How many
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Answer:

3,000 meters

Step-by-step explanation:

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the value of c guaranteed to exist by the Mean Value Theorem for V(x) = x² in the interval [0, 3] is...? a) 1 b) 2 c) 3/2 d) 1/2
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Mean value theorem : If f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that

\begin{displaymath}f'(c) = \frac{f(b) - f(a)}{b-a} \cdot\end{displaymath}

Given function : f(x) = x^2

Interval : [0,3]

Then, by the mean value theorem, there is at least one number c in the interval (0,3) such that

f'(c)=\dfrac{f(3)-f(0)}{3-0}\\\\=\dfrac{3^2-0^2}{3}=\dfrac{9}{3}\\\\=3

\Rightarrow\ f'(c)=3\ \ \ ...(i)

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From (i) and (ii), we have

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