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2 years ago

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Vektor i adalah vektor satuan yang searah dengan sumbu X (dalam dimensi dua). Vektor jika dinyatakan dalam bentuk kolom adalah..

..

1 answer:

sineoko [7]2 years ago

5 0

Answer:

$v=\left[\begin{array}{ccc}1i\\0j\end{array}\right]$

Step-by-step explanation:

We know that the given unit vector is in the direction of the x-axis.

Remember that an unit vector is a vector which magnitude is the unit, that's why is called like that.

So, the given unit vector can be written as $v=1i+oj$ , where $i$ indicates the horziontal coordinate and $j$ indicates the vertical coordinate.

As a vertical or colum vector would be

$v=\left[\begin{array}{ccc}1i\\0j\end{array}\right]$

Therefore, the answer is

$v=\left[\begin{array}{ccc}1i\\0j\end{array}\right]$

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Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?

irakobra [83]

First, note that $m\angle A+m\angle C=90^{\circ}.$ Then

$m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.$

Consider all options:

A.

$\tan A=\dfrac{\sin A}{\sin C}$

By the definition,

$\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.$

Now

$\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.$

Option A is true.

B.

$\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.$

By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.$

Then

$\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.$

Option B is false.

3.

$\sin C = \dfrac{\cos A}{\tan C}.$

By the definition,

$\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

Now

$\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.$

Option C is false.

D.

$\cos A=\tan C.$

By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

As you can see $\cos A\neq \tan C$ and option D is not true.

E.

$\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.$

By the definition,

$\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.$

Then

$\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.$

This option is false.

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2 years ago

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Andros has an oil fired heating system. In a 30-day period he used a full tank of oil at a constant rate per day. At a different

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Step-by-step explanation:

30 days period multiply by 7 days

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A triangle on a coordinate plane is translated according to the rule T-3, 5(x, y). Which is another way to write this rule?

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Answer:c

Step-by-step explanation:

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Two boats depart from a port located at (–8, 1) in a coordinate system measured in kilometers and travel in a positive x-directi

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Answer:

$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

Step-by-step explanation:

1st boat:

Parabola equation:

$y=ax^2 +bx+c$

The x-coordinate of the vertex:

$x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=1\\ \\b=-2a$

Equation:

$y=ax^2 -2ax+c$

The y-coordinate of the vertex:

$y_v=a\cdot 1^2-2a\cdot 1+c\Rightarrow a-2a+c=10\\ \\c-a=10$

Parabola passes through the point (-8,1), so

$1=a\cdot (-8)^2-2a\cdot (-8)+c\\ \\80a+c=1$

Solve:

$c=10+a\\ \\80a+10+a=1\\ \\81a=-9\\ \\a=-\dfrac{1}{9}\\ \\b=-2a=\dfrac{2}{9}\\ \\c=10-\dfrac{1}{9}=\dfrac{89}{9}$

Parabola equation:

$y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}$

2nd boat:

Parabola equation:

$y=ax^2 +bx+c$

The x-coordinate of the vertex:

$x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=0\\ \\b=0$

Equation:

$y=ax^2+c$

The y-coordinate of the vertex:

$y_v=a\cdot 0^2+c\Rightarrow c=-7$

Parabola passes through the point (-8,1), so

$1=a\cdot (-8)^2-7\\ \\64a-7=1$

Solve:

$a=-\dfrac{1}{8}\\ \\b=0\\ \\c=-7$

Parabola equation:

$y=\dfrac{1}{8}x^2 -7$

System of two equations:

$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

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Answer:

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Step-by-step explanation:

p(d) = 24p + 500d

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Can you answer my question pleaseeee?

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