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3 years ago

There are 90 monkeys in a zoo.

Mathematics

2 answers:

Vsevolod [243]3 years ago

8 0

I think it would be 103

KonstantinChe [14]3 years ago

4 0

Answer:

Step-by-step explanation:

Well funny enough I think this is a twisted quiz coz directly the last statement says;24 like to eat both apples and bananas,so those that don't like both but like either of them is simply 90-24=66.

My thinking ,oops

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24+3/8w. W=1/2. I need help nowwwww

aleksklad [387]

Answer:

24 3/16

Step-by-step explanation:

To work out the question, you first have to replace the letter with its numerical value. In this case, W = ½

therefore it becomes,

24 + (⅜ ✖ ½)

24 + 3/16

= 24 3/16

I hope it helps

4 0

3 years ago

I need help boi a(c – b) = d c =

alexandr1967 [171]

Hello from MrBillDoesMath!

Answer:

c = d/a + b

Discussion:

I think the Question is missing a comma. Assuming you are trying to find the value "c",

a(c-b) = d => divide both side by "a"

(c-b) = d/a => add "b" to both sides

c-b + b = d/a + b => as -b + b = 0

c = d/a + b

Thank you,

MrB

5 0

3 years ago

MArishka [77]

Answer:

b. angle I

Step-by-step explanation:

3 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.