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3 years ago

What is cos(45°)?

Mathematics

1 answer:

Andrej [43]3 years ago

6 0

Answer:

$\rm \dfrac{\sqrt2}{2}$

Step-by-step explanation:

A right angled ∆ is given to us and we need to find out the value of cos 45° ( with rationalized denominator ) . As we know that in a right angle ∆ , the value of cosine is ,

$\rm\implies cos\theta =\dfrac{base}{hypotenuse}$

In the given triangle ,

Value of base = 1
Value of hypotenuse = √2

On substituting the respective values ,

$\rm\implies cos45^\circ =\dfrac{1}{\sqrt2}$

For rationalizing the denominator , multiply numerator and denominator by √2 , we have,

$\rm\implies cos45^\circ=\dfrac{1(\sqrt2)}{(\sqrt2)(\sqrt2)}$

Simplify ,

$\rm\implies \boxed{\pink{\frak{ cos45^\circ=\dfrac{\sqrt2}{2} }}}$

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galina1969 [7]

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3 years ago

Read 2 more answers

Eggs flour and sugar are mixed in the ratio of 3: 5: 4 by mass to make a cake. If the mass of the sugar is 240g find the mass of

zhuklara [117]

Answer:

eggs four sugar

3 5 4

mass of eggs;

4=240

3=?

3x240/4

mass of eggs=180g

mass of flour

4=240

5=?

5x240/4

mass of flour =300g

Step-by-step explanation:

5 0

2 years ago

Hi need help for this maths question

Sauron [17]

a) If f(y) is a probability density function, then both f(y) ≥ 0 for all y in its support, and the integral of f(y) over its entire support should be 1. eˣ > 0 for all real x, so the first condition is met. We have

$\displaystyle \int_{-\infty}^\infty f(y) \, dy = \frac14 \int_0^\infty e^{-\frac y4} \, dy = -\left(\lim_{y\to\infty}e^{-\frac y4} - e^0\right) = \boxed{1}$

so both conditions are met and f(y) is indeed a PDF.

b) The probability P(Y > 4) is given by the integral,

$\displaystyle \int_{-\infty}^4 f(y) \, dy = \frac14 \int_0^4 e^{-\frac y4} \, dy = -\left(e^{-1} - e^0\right) = \frac{e - 1}{e} \approx \boxed{0.632}$

c) The mean is given by the integral,

$\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \int_0^\infty y e^{-\frac y4} \, dy$

Integrate by parts, with

$u = y \implies du = dy$

$dv = e^{-\frac y4} \, dy \implies v = -4 e^{-\frac y4}$

Then

$\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \left(\left(\lim_{y\to\infty}\left(-4y e^{-\frac y4}\right) - \left(-4\cdot0\cdot e^0\right)\right) + 4 \int_0^\infty e^{-\frac y4} \, dy\right)$