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mr Goodwill [35]
3 years ago
11

Please give the answer! thank you!! <3 :D​

Mathematics
2 answers:
DIA [1.3K]3 years ago
7 0
8x is the answer to this question
Eduardwww [97]3 years ago
4 0

Answer:

8x Should be the answer to the question

You might be interested in
If the radius of the circle is 7.2cm what is the circumference to the nearest tenth?
Aleksandr-060686 [28]

Answer:

About 45.22

Step-by-step explanation:

Radius x 2 x π

8 0
3 years ago
I need help with this problem please! It's very confusing to me.
Daniel [21]
2.25X5= 11.25

2 1/4 turns into 2.25 as a decimal, and a pentagon has 5 equal sides, so you take 2.25 and multiply it by 5 to get your answer

11.25ft
4 0
3 years ago
Solve the question below
Tatiana [17]

Answer:

AAS Congruence Theorem

Step-by-step explanation:

Please let me know if you want me to add an explanation as to why this is the answer. I can definitely do that, I just don’t want to waste my time in case you don’t want me to :)

7 0
3 years ago
What is the value of the expression below. <br> ||3-7| - |-6+3||
alexdok [17]

Answer:

1

Step-by-step explanation:

||3-7| - |-6+3||

||-4|- |-3||

|4-3|

|1|

6 0
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}}
3 0
4 years ago
Read 2 more answers
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