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3 years ago

In the diagram of circle o, what is the measure of ZABC? O 30° O 40° O 50° O 60

Mathematics

2 answers:

denpristay [2]3 years ago

7 0

Answer:

30°

Step-by-step explanation:

Line AB and BC are tangents to the given circle.

$\angle \: ABC = \frac{1}{2} ( 210 - 150)$

$\angle \: ABC = \frac{1}{2} (60) = 30 \degree$

Alternatively, <ABC and <AOC are supplementary because AB and BC are tangents.

$\angle \: ABC + 150 \degree = 180 \degree$

$\angle \: ABC = 180 \degree - 150 \degree = 30 \degree$

The correct choice is A.

netineya [11]3 years ago

3 0

Answer:30

Step-by-step explanation:

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Step-by-step explanation:

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gizmo_the_mogwai [7]

We can rewrite the squared in two ways.

The first way is to rewrite it as a multiplication, this is:

$(y-6)^2=(y-6)(y-6)$

The second way to rewrite the squared is by using the formula for this kind of product:

$(y-6)^2=y^2+2(-6)(y)+(-6)^2$

Once again we can find the final result using each of the options given in part A, for the first option we have:

$\begin{gathered} (y-6)^2=(y-6)(y-6) \\ =y^2-6y-6y+36 \\ =y^2-12y+36 \end{gathered}$

For the second option we have:

$\begin{gathered} (y-6)^2=y^2+2(-6)(y)+(-6)^2 \\ =y^2-12y+36 \end{gathered}$

No matter which way we choose the answer for the squared is:

$y^2-12y+36$

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1 year ago

tim and alex collected aluminum cans for recycling. Tim collected a total of 942 cans. Alex collected 327 cans. How many fewer c

lukranit [14]

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3 years ago

Read 2 more answers

Can someone help me out ! got stuck in this question for an hour.

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

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

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3 years ago