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3 years ago

I need to know what’s 500+0+12+44+55+500+0+12+44+55+ 500+0+12+44+55+500+0+12+44+55+

Mathematics

2 answers:

ehidna [41]3 years ago

4 0

Answer:

2,444.

Step-by-step explanation:

500+0+12+44+55+500+0+12+44+55+ 500+0+12+44+55+500+0+12+44+55 = 500 + 12 + 44 + 55 + 500 + 12 + 44 + 55 + 500 + 12 + 44 + 55 + 500 + 12 + 44 + 55 = 4(500 + 12 + 44 + 55) = 4(512 + 99) = 4(611) = 2,444.

Hope this helps!

Levart [38]3 years ago

4 0

Answer:

Step-by-step explanation:

2,444

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

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

$b = 3$

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2 years ago

Someone please help me on this and show work I need help I’ll give brainliest

irinina [24]

A.

F(10)= (10)^2-3(10)-5
F(10)=100-30-5
F(10)=65

B.

F(-3)=(-3)^2-3(-3)-5
F(-3)=9+9-5
F(-3)=13

C.

G(2)= -6(2)+1
G(2)= -12+1
G(2)=-11

D.

G(10)=-6(10)+1
G(10)= -60+1
G(10)= -59

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3 years ago

The class is 60% women and 10% of the women have blond hair, so blond women comprise 60% × 10% = 6% of the class.

Tatiana [17]

Answer:

Conversion and substitution

Step-by-step explanation:

One way to find out if the expression is true, we can convert all the percentages into their decimal equivalence.

60% = 0.6

10% = 0.1

6% = 0.06

Now let's substitute the values.

0.06 = 0.6 x 0.1

0.06 = 0.06

Since 0.06 is equal 0.06, this then proves that it is true.

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3 years ago