Questions: 29. vec(a)+vec(b)=(1 ; 3 ; 2), overrightarrow(2 a)+vec(b)=(3 ; 4 ; 1) баих a ба b векторуудын хоорондох өнщгиин косинусыг олоорой. A. (2 sqrt(21))/2 B. -(sqrt(21))/14 C. -(2 sqrt(17))/17 D. -(7)/(3 sqrt(15)) E. (sqrt(55))/11 30. 2 ажилчин хамтран 2 өдөрг ажлаа дуустав. Хэрэв I ажилчин нь 2 өдөр, II ажилчин пь 1 өдөр ажилласан бол тэд, нийллээд, нийт ажлын 5/6-ийг гүйцэтгэх байв. I ажилчин ганцаараа энэхүү ажлын хэдэн өдөр гүйцэтгэх вэ? A. 1 B. 2 C. 3 D. 4 E. 5 32. int-1^2 2(t-1) x d x=2 t бол t= ? A. -2 B. -3 C. 2 D. 3 E. 0

Solution Steps

Given: \[ \vec{a} + \vec{b} = (1, 3, 2) \] \[ 2\vec{a} + \vec{b} = (3, 4, 1) \]

First, subtract the first equation from the second: \[ (2\vec{a} + \vec{b}) - (\vec{a} + \vec{b}) = (3, 4, 1) - (1, 3, 2) \] \[ \vec{a} = (2, 1, -1) \]

Now substitute \(\vec{a}\) back into the first equation to find \(\vec{b}\): \[ (2, 1, -1) + \vec{b} = (1, 3, 2) \] \[ \vec{b} = (1, 2, 3) \]

Next, find the dot product \(\vec{a} \cdot \vec{b}\): \[ \vec{a} \cdot \vec{b} = 2 \cdot 1 + 1 \cdot 2 + (-1) \cdot 3 = 2 + 2 - 3 = 1 \]

Find the magnitudes of \(\vec{a}\) and \(\vec{b}\): \[ |\vec{a}| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \] \[ |\vec{b}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

The cosine of the angle between \(\vec{a}\) and \(\vec{b}\) is: \[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{1}{\sqrt{6} \cdot \sqrt{14}} = \frac{1}{\sqrt{84}} = \frac{1}{2\sqrt{21}} \]

Thus, the answer is: \[ \boxed{\frac{\sqrt{21}}{42}} \]

Let \(x\) be the number of days it takes for the first worker to complete the job alone, and \(y\) be the number of days it takes for the second worker to complete the job alone.

Given: \[ \frac{2}{x} + \frac{2}{y} = 1 \] \[ \frac{2}{x} + \frac{1}{y} = \frac{5}{6} \]

First, solve the first equation for \(\frac{2}{y}\): \[ \frac{2}{y} = 1 - \frac{2}{x} \]

Substitute into the second equation: \[ \frac{2}{x} + \frac{1}{1 - \frac{2}{x}} = \frac{5}{6} \]

Solve for \(x\): \[ \frac{2}{x} + \frac{1}{1 - \frac{2}{x}} = \frac{5}{6} \]

This is a complex equation, so let's simplify it by solving the system of equations directly. Multiply the first equation by 3 and the second by 2: \[ 3 \left( \frac{2}{x} + \frac{2}{y} \right) = 3 \cdot 1 \implies \frac{6}{x} + \frac{6}{y} = 3 \] \[ 2 \left( \frac{2}{x} + \frac{1}{y} \right) = 2 \cdot \frac{5}{6} \implies \frac{4}{x} + \frac{2}{y} = \frac{5}{3} \]

Subtract the second equation from the first: \[ \left( \frac{6}{x} + \frac{6}{y} \right) - \left( \frac{4}{x} + \frac{2}{y} \right) = 3 - \frac{5}{3} \] \[ \frac{2}{x} + \frac{4}{y} = \frac{4}{3} \]

Now solve for \(x\): \[ \frac{2}{x} = \frac{4}{3} - \frac{4}{y} \]

Substitute back into the first equation: \[ \frac{2}{x} + \frac{2}{y} = 1 \]

After solving, we find: \[ x = 3 \]

Thus, the answer is: \[ \boxed{3} \]

Given the interval \(10 ; 4]\), we need to find the correct interval representation.

The correct interval representation is: \[ \boxed{[0 ; 1]} \]

Final Answer