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Sauron [17]
3 years ago
10

9+7+(+^8+7+9 and 12 more

Mathematics
2 answers:
MariettaO [177]3 years ago
7 0
The answer will be 54 
Soloha48 [4]3 years ago
3 0
The answer will be 52. hope it helps
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What two steps are necessary to put this equation into standard form?
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X² - 3x + 27 = 8x - 3     аdd 3 to both sides  ⇒
           + 3          + 3
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x² - 3x + 30 = 8x         subtract 8x from both sides ⇒<span>
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x</span>² - 11x + 30 = 0    ←  <span>standard form
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A. Add 3 to both sides and subtract 8x from both sides.
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3 years ago
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In a hostel, 120 students had provisions of food for 60 days. After 20 days, some students were added to the hostel, the provisi
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If I'm right, the answers should be 40.

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2 years ago
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Dude, that was not nice he just wants to know!

Step-by-step explanation:


6 0
3 years ago
Find the angle between u =the square root of 5i-8j and v =the square root of 5i+j.
fenix001 [56]

Answer:

The angle between vector \vec{u} = 5\, \vec{i} - 8\, \vec{j} and \vec{v} = 5\, \vec{i} + \, \vec{j} is approximately 1.21 radians, which is equivalent to approximately 69.3^\circ.

Step-by-step explanation:

The angle between two vectors can be found from the ratio between:

  • their dot products, and
  • the product of their lengths.

To be precise, if \theta denotes the angle between \vec{u} and \vec{v} (assume that 0^\circ \le \theta < 180^\circ or equivalently 0 \le \theta < \pi,) then:

\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}.

<h3>Dot product of the two vectors</h3>

The first component of \vec{u} is 5 and the first component of \vec{v} is also

The second component of \vec{u} is (-8) while the second component of \vec{v} is 1. The product of these two second components is (-8) \times 1= (-8).

The dot product of \vec{u} and \vec{v} will thus be:

\begin{aligned} \vec{u} \cdot \vec{v} = 5 \times 5 + (-8) \times1 = 17 \end{aligned}.

<h3>Lengths of the two vectors</h3>

Apply the Pythagorean Theorem to both \vec{u} and \vec{v}:

  • \| u \| = \sqrt{5^2 + (-8)^2} = \sqrt{89}.
  • \| v \| = \sqrt{5^2 + 1^2} = \sqrt{26}.

<h3>Angle between the two vectors</h3>

Let \theta represent the angle between \vec{u} and \vec{v}. Apply the formula\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} to find the cosine of this angle:

\begin{aligned} \cos(\theta)&= \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} = \frac{17}{\sqrt{89}\cdot \sqrt{26}}\end{aligned}.

Since \theta is the angle between two vectors, its value should be between 0\; \rm radians and \pi \; \rm radians (0^\circ and 180^\circ.) That is: 0 \le \theta < \pi and 0^\circ \le \theta < 180^\circ. Apply the arccosine function (the inverse of the cosine function) to find the value of \theta:

\displaystyle \cos^{-1}\left(\frac{17}{\sqrt{89}\cdot \sqrt{26}}\right) \approx 1.21 \;\rm radians \approx 69.3^\circ .

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3 years ago
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3 years ago
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