3 4/7 minus 2 1/5 by rounding to the nearest half

Mathematics

1 answer:

mylen [45]3 years ago

6 0

Answer:

$1 \frac{13}{35}$

Step-by-step explanation:

$3 \frac{4}{7} - 2 \frac{1}{5}$

you can solve for the fractions first, but you have to give them a common denominator

$\frac{20}{35}- \frac{7}{35}$

$= \frac{13}{35}$ which is in simplest form

then solve for the whole numbers

3-2=1

therefore, the answer is $1 \frac{13}{35}$

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3 years ago

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

Oksi-84 [34.3K]

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: $...\\\\ E-1=(2^{32}-1)(2^{32}+1)\\\\ E-1=2^{64}-1\\\\ \boxed{E=2^{64}}$