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3 years ago

.please help me with this I will crown you brainlest

Mathematics

2 answers:

Brut [27]3 years ago

8 0

I think it is either A or C
but more sure about A

Marizza181 [45]3 years ago

4 0

You can prove it by the SAS postulate which should be the first option

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Givem the geometric sequence where à1=2 and the common ratio is 4, what is domain for N

Dmitry_Shevchenko [17]

The domain for N is All integers where n ≥ 1

<u>Solution:</u>

According to statement a1 = 2 and r = 4. This shows that r is greater than 1.

If r is greater than 1 than it includes integers greater than 1 or equal to 1. It does not include all the real numbers because real numbers include negative numbers also.

If starting value is 2, if we put n=0, then we get 2, but if we put a negative value than we would get a number which is not a part of our sequence.

Thus the domain of n is All integers where n greater than or equal to 1

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4 years ago

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kompoz [17]

SA=240sq.cm

have an amazing day <3

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3 years ago

At the fundraiser, Chad washed 7 cars every 3 hours and earned over $150 for the school dance. 3. Choose the correct answer. Whi

Vladimir [108]

Answer:

the answer is 22

Step-by-step explanation:

4 0

3 years ago

Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.