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qaws [65]
3 years ago
5

Write the smallest possible odd number with all the following dight 5,2,3,7.​

Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
5 0

Answer:

2357

Step-by-step explanation:

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Simplify the expression <br>16+(-5)-2/5j-4/5j+6​
stiks02 [169]

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17 - 6j / 5

Step-by-step explanation:

8 0
3 years ago
Can someone help me out !<br><br>got stuck in this question for an hour.<br><br><br>​
Reil [10]

Answer:

See below

Step-by-step explanation:

Considering $\vec{u}, \vec{v}, \vec{w} \in V^3 \lambda \in \mathbb{R}$, then

\Vert \vec{u} \cdot \vec{v}\Vert \leq  \Vert\vec{u}\Vert  \Vert\vec{v}\Vert$ we have $(\vec{u} \cdot \vec{v})^2 \leq (\vec{u} \cdot \vec{u})(\vec{v} \cdot \vec{v}) \quad$

This is the Cauchy–Schwarz  Inequality, therefore

$\left(\sum_{i=1}^{n} u_i v_i \right)^2 \leq \left(\sum_{i=1}^{n} u_i \right)^2 \left(\sum_{i=1}^{n} v_i \right)^2  $

We have the equation

\dfrac{\sin ^4 x }{a} + \dfrac{\cos^4 x }{b}  = \dfrac{1}{a+b}, a,b\in\mathbb{N}

We can use the Cauchy–Schwarz  Inequality because a and b are greater than 0. In fact, a>0 \wedge b>0 \implies ab>0. Using the Cauchy–Schwarz  Inequality, we have

\dfrac{\sin ^4 x }{a} + \dfrac{\cos^4 x }{b}   =\dfrac{(\sin^2 x)^2}{a}+\dfrac{(\cos^2 x)}{b}\geq \dfrac{(\sin^2 x+\cos^2 x)^2}{a+b} = \dfrac{1}{a+b}

and the equation holds for

\dfrac{\sin^2{x}}{a}=\dfrac{\cos^2{x}}{b}=\dfrac{1}{a+b}

\implies\quad \sin^2 x = \dfrac{a}{a+b} \text{ and }\cos^2 x = \dfrac{b}{a+b}

Therefore, once we can write

\sin^2 x = \dfrac{a}{a+b} \implies \sin^{4n}x = \dfrac{a^{2n}}{(a+b)^{2n}} \implies\dfrac{\sin^{4n}x }{a^{2n-1}} = \dfrac{a^{2n}}{(a+b)^{2n}\cdot a^{2n-1}}

It is the same thing for cosine, thus

\cos^2 x = \dfrac{b}{a+b} \implies \dfrac{\cos^{4n}x }{b^{2n-1}} = \dfrac{b^{2n}}{(a+b)^{2n}\cdot b^{2n-1}}

Once

\dfrac{a^{2n}}{(a+b)^{2n}\cdot a^{2n-1}}+ \dfrac{b^{2n}}{(a+b)^{2n}\cdot b^{2n-1}} =\dfrac{a^{2n}}{(a+b)^{2n} \cdot \dfrac{a^{2n}}{a} } + \dfrac{b^{2n}}{(a+b)^{2n}\cdot \dfrac{b^{2n}}{b} }

=\dfrac{1}{(a+b)^{2n} \cdot \dfrac{1}{a} } + \dfrac{1}{(a+b)^{2n}\cdot \dfrac{1}{b} } = \dfrac{a}{(a+b)^{2n}  } + \dfrac{b}{(a+b)^{2n} } = \dfrac{a+b}{(a+b)^{2n} }

dividing both numerator and denominator by (a+b), we get

\dfrac{a+b}{(a+b)^{2n} } =  \dfrac{1}{(a+b)^{2n-1} }

Therefore, it is proved that

\dfrac{\sin ^{4n} x }{a^{2n-1}} + \dfrac{\cos^{4n} x }{b^{2n-1}}  = \dfrac{1}{(a+b)^{2n-1}}, a,b\in\mathbb{N}

4 0
3 years ago
A rectangular garden has length and width as given by the expressions below. Length: 4 − 7(3x + 4y) Width: 3x(−2y) Write a simpl
liraira [26]

Answer:

Step-by-step explanation:

Length of the garden = 4 - 21*x - 28*y

Width of the garden = -6*x*y

Perimeter of the garden = 2*(Length + Width)

P = 8 - 42*x - 56*y - 12*x*y

4 0
3 years ago
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