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Llana [10]
3 years ago
8

Anyone knows how to do questions 7 and 8? 15 pts!!

Mathematics
2 answers:
Oksi-84 [34.3K]3 years ago
3 0
7) Certainly there is a typo in the statement, just see that the expression of item (ii) is different from that of item (i). Probably the correct expression is: 2x^2-4x+5. With this consideration, we can continue.

(i) Let E the expression that we are analyzing:

E=2x^2-4x+5\\\\ E=2x^2-4x+2-2+5\\\\ E=2(x^2-2x+1)-2+5\\\\ E=2(x-1)^2+3

Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.

(ii) Manipulating the expression:

2x^2+5=4x\\\\ 2x^2-4x+5=0

So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.

8) Let E the expression that we want to calculate:

E=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)+1\\\\ E-1=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)

Multiplying by (2-1) in the both sides:

(2-1)(E-1)=(2-1)(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2-1)(2+1)}_{2^2-1}(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2^2-1)(2^2+1)}_{2^4-1}(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ ... Repeating the process, we obtain: ...\\\\ E-1=(2^{32}-1)(2^{32}+1)\\\\ E-1=2^{64}-1\\\\ \boxed{E=2^{64}}
kogti [31]3 years ago
3 0
Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.
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133,7697

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A certain bookstore chain has two stores, one in San Francisco and one in Los Angeles. It stocks three kinds of books: hardcover
riadik2000 [5.3K]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

(a)

\left[\begin{array}{cccc}T&H&S&P\\S&600&1300&2000\\L&400&300&400\end{array}\right] = A.

In the above matrix A, the columns refers the three type of books and the rows refers the from which stores the books are been sold.

The numbers represents the corresponding sales in the month of January.

The sale is same for the 6 months.

Hence, 6A = \left[\begin{array}{cccc}T&H&S&P\\S&3600&7800&12000\\L&2400&1800&2400\end{array}\right]. This matrix 6A represents the total sales over the 6 months.

(b)

If we denote the books in stock at the starting of January by B, then

B = \left[\begin{array}{cccc}T&H&S&P\\S&1000&3000&6000\\L&1000&6000&3000\end{array}\right].

Each month, the chain restocked the stores from its warehouse by shipping 500 hardcover, 1,400 softcover, and 1,400 plastic books to San Francisco and 500 hardcover, 500 softcover, and 500 plastic books to Los Angeles.

If we represent the amount restocked books at the end of each month by another matrix C, then

C = \left[\begin{array}{cccc}T&H&S&P\\S&500&1400&1400\\L&500&500&500\end{array}\right].

This restocking will be done for 5 times before the end of June.

If there would be no sale, then the stock would be

B + 5C = \left[\begin{array}{cccc}T&H&S&P\\S&1000+2500&3000+7000&6000+7000\\L&1000+2500&6000+2500&3000+2500\end{array}\right] \\= \left[\begin{array}{cccc}T&H&S&P\\S&3500&10000&13000\\L&3500&8500&5500\end{array}\right].

Since, the total sale is given by 6A, at the end of June, the inventory in each store can be shown as following,

B+5C-6A \left[\begin{array}{cccc}T&H&S&P\\S&3500&10000&13000\\L&3500&8500&5500\end{array}\right] - \left[\begin{array}{cccc}T&H&S&P\\S&3600&7800&12000\\L&2400&1800&2400\end{array}\right] \\= \left[\begin{array}{cccc}T&H&S&P\\S&-100&2200&1000\\L&1100&6700&3100\end{array}\right]

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3 years ago
How to solve it and the formula to solve it
notsponge [240]

i) u + 3 V

1 by 2 + 3 *minus 2

1 by 2 -6

1 - 12 by 2

- 11 by 2

ii) - 3 * 1 by 2 + 2 * -2 + 3 by 4

minus 19 by 4

7 0
3 years ago
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