How do you transform the polar equation theta=pi/6 to an equation in rectangular coordinates and identify its shape?

1 answer:

5 0

<span>You are given the theta function </span>θ=π/6, suppose that you have a radius of 4, then we have a polar coordinate of (4,π/6). To change into rectangular form, this is also equal to
x = 4cosπ/6 and y = 4sinπ/6
x = 2√3 and y = 2
(2√3, 2)

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Solve for t. d= -16t^2 + 12

IRINA_888 [86]

Answer:

t=√d-12/4 and t=√-d-12/4, d≥12

Step-by-step explanation:

Hope this helps.

3 0

2 years ago

monica tiene 6/10kg de dulces con los cuales llena 1/4 de una bolsa de celofn ¿cual es la capacidad de la bolsa?

Anit [1.1K]

The question in English
<span>Monica has 6 /10 kg of candy with which she fills 1/4 of a cellophane bag. What is the capacity of the bag?
</span>
we know that
6/10---------> 0.60
1/4----------> 0.25------> 25%

if 25% of a cellophane bag---------------> has a 0.60 kg of candy
100%-----------------------------------------> X
X=100*0.60/25--------> X=2.4 kg

the answer is
2.4 kg

the answer in Spanish

Sabemos que
6/10---------> 0.60
1/4----------> 0.25------> 25%

if 25% de la bolsa de celofan---------------> se llena con 0.60 kg de dulces
100% de la bolsa----------------------------------------->se llenara con X kg
X=100*0.60/25--------> X=2.4 kg

La respuesta es
2.4 kg

5 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.