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3 years ago

2+2=? i need help i don't know the answer

Mathematics

2 answers:

charle [14.2K]3 years ago

8 0

Answer:

It is four

Step-by-step explanation:

When you add 2+2 always put 2 in you head and count 2 3 4 and the you say 2+2=4 that's why the answer is four

Dmitriy789 [7]3 years ago

7 0

Step-by-step explanation: First you put two pencil then add another two pencil. Now how much do you have in total, the answer is 4.

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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.

In addition, $b$ must be a natural number, which means that:

$1 + 2\cdot a\cdot c \ge 4$

$2\cdot a \cdot c \ge 3$

$a\cdot c \ge \frac{3}{2}$

But the lowest possible product made by two prime numbers is $2^{2} = 4$ . Hence, $a\cdot c \ge 4$ .

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)

Example: $a = 2$ , $c = 2$

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

$b = 3$

4 0

3 years ago

Which statement is true about the domain of y = 3(2^-x)? Explain.

SVEN [57.7K]

We want to determine the domain of ${y=3 \cdot 2^{-x}=3 \cdot ({2^{-1}})^x=3 \cdot ({ \frac{1}{2}})^x$

any function of the form $y=f(x)=a \cdot b^x$ is called an "exponential function",
the only condition is that b is positive and different from 1, and a is a nonzero real number.

The domain of such functions is all real numbers.

That is for any x, the expression 3(2^-x) "makes sense".

Answer: The domain is all real numbers

8 0

3 years ago

A television game has 6 shows doors, of which the contests must pick 2. behind two of the doors are expensive cars, and behind t

Katyanochek1 [597]

The answer to this question:
One car probability 82/120
No car probability = 24/120
At least one car probability= 96/120

I will focus answering the 3 doors probability since the 2nd door problem is solved in the previous problem. (brainly.com/question/5761449)

No car condition
1. 1st door consolation, 2nd door consolation=, 3rd door consolation= 4/6 * 3/5 * 2/4= 24/120
This was also can be found by: (4!/1!)/ (6!/3!) = 24/120

(At least one car probability) is the opposite of (no car probability) In this case, the easier way is
100% - (no car probability) = 120/120 - 24/120= 96/120

One car probability is (At least one car probability) - (2 car probability). It will be easier to count the 2 car probability and subtract the (At least one car probability)
Two car condition:
1. 1st door car, 2nd door car, 3rd door consolation = 2/6 * 1/5 * 4/4 =8/ 120
2.1st door car, 2nd door consolation, 3rd door car =2/6 * 4/5 * 1/4 = 8/120
3. 1st door consolation, 2nd door car, 3rd door car= 4/6 * 2/5 * 1/4= 8/120
The total probability will be 8/120+ 8/120 + /120= 24/120
This was also can be found by: (2!) (4!/2!)/ (6!/3!) = 24/120

One car probability = (At least one car probability) - (2 car probability)= 96/120-24/120= 82/120

3 0

3 years ago

Solve for the variable z to 4 = 8/5

Shkiper50 [21]

Answer:

Alright well solve for z by simplifying by both sides of the equation, then isolating the variable

Exact form: z = 2/5

Decimal form: o.4 Hope this helps :)

Step-by-step explanation:

5 0

3 years ago

How many real solutions does the equation V2x + 4 = 3x + 1 have?

Verdich [7]

Answer:

5v^2x + 4

Step-by-step explanation:

I'm not sure but i hope this help's

3 0

3 years ago