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iren2701 [21]
3 years ago
13

If sin(x)=m, with m>0, find the value of sin(pie-x)

Mathematics
1 answer:
viva [34]3 years ago
8 0

Answer:

Step-by-step explanation:

Sin (π - x ) = Sin x = m

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6/13=3/c what I the answer and show me how you got your answer.show me how you did it.
umka21 [38]

          6             3
        ----      *    ----
        13              c

this is a proportion...so we cross multiply
(6)(c) = (3)(13)
6c = 39...divide both sides by 6
c = 39/6 which reduces to 13/2 or 6 1/2 <== ur answer
4 0
3 years ago
Who was born in California and makes use of symbolism, folklore, oral tradition, popular stories and songs within his novels?
VladimirAG [237]
The correct answer is John Steinbeck. He is famous for his depiction of Californian common folk lives and even won a Nobel prize for his depictions of it.
4 0
2 years ago
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What is the area of a half circle with a radius of 40 cm?
Andre45 [30]

Answer:

5027 square centimeters (cm2)

Step-by-step explanation:

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2 years ago
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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
What value of k makes the factor (x+3) a factor of the function f(x)=3x^3-2x+k?
4vir4ik [10]
Step by step equation
5 0
3 years ago
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