Ask question

Ask question

All categories

3 years ago

5

If P(x, y) is the point on the unit circle defined by real number 0, then cot 0=

1 answer:

denpristay [2]3 years ago

4 0

Answer:

x/y

Step-by-step explanation:

The question is badly worded, to the point of being impossible to answer as it is written.

We presume the given point (x, y) is (cos(θ), sin(θ)). The definition of cot(θ) is cos(θ)/sin(θ), so matches x/y with our assumed definitions of x and y.

You might be interested in

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

5 0

3 years ago

Solve exponents please help just this one u get 10 points

STALIN [3.7K]

I think it's equal to 1 because every expression with the exponent of zero is equal to 1, idk for sure though

8 0

3 years ago

Y = -2 x - 5 what is the answer help

blondinia [14]

Answer:

x = -5/2

Step-by-step explanation:

0 = -2x - 5

2x = -5

Divide^

x = -5/2

Hope this helped

Brainliest is appreicated./

7 0

3 years ago

2x^2+7x=15 write in standard then factor the left side of the equation.

Nataly [62]

Answer:

see explanation

Step-by-step explanation:

Given

2x² + 7x = 15 ( subtract 15 from both sides )

2x² + 7x - 15 = 0 ← in standard form

To factorise the left side

Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term

product = 2 × - 15 = - 30 and sum = + 7

The factors are + 10 and - 3

Use these factors to split the x- term

2x² + 10x - 3x - 15 ( factor the first/second and third/fourth terms )

2x(x + 5) - 3(x + 5) ← factor out (x + 5) from each term

(x + 5)(2x - 3) ← in factored form

5 0

3 years ago

This question is not that hard but I’m getting rushed

Vsevolod [243]

14:21 one is the right answer

4 0

3 years ago