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4 years ago

Find the angle between vector u=2i+sqrt(11)j and v=-3i-2j to the nearest degree

Mathematics

1 answer:

const2013 [10]4 years ago

7 0

Answer: Last option 155°

Step-by-step explanation:

We have the components of the vectors u and v.

Then, to find the angle between them, perform the following steps:

1) Calculate the scalar product u * v

If $u = 2i + \sqrt{11}j$ and $v = -3i-2j$

Then, the product scalar u * v is:

$u * v = (2)(-3) + (\sqrt{11})(-2)$

$u * v = -6-2\sqrt{11}$

$u * v = -12.633$

2) Calculation of the magnitude of both vectors.

$| u | = \sqrt{(2) ^ 2 + (\sqrt{11})^2}\\\\| u | = \sqrt{4+11}\\\\| u | = \sqrt{15}$

$| v | = \sqrt{(- 3) ^ 2 + (- 2) ^ 2}\\\\| v | = \sqrt{9 +4}\\\\| v | = \sqrt{13}$

3) Now that you know the product point between the two vectors and the magnitude of each, then use the following formula to find an angle

$u * v = | u || v |cos(\alpha)$

$-12.633 = \sqrt{15}\sqrt{13}*cos(\alpha)\\\\cos(\alpha) = \frac{-12.633}{\sqrt{15}\sqrt{13}}\\\\arcos(\frac{-12.633}{\sqrt{15}\sqrt{13}}) = \alpha\\\\\alpha =155\°$

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Ilya [14]

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3 0

3 years ago

Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = t 25 − t2 , [−1, 5]

Zina [86]

Answer: Absolute minimum: f(-1) = -2 $\sqrt{6}$

Absolute maximum: f( $\sqrt{12.5}$ ) = 12.5

Step-by-step explanation: To determine minimum and maximum values in a function, take the first derivative of it and then calculate the points this new function equals 0:

f(t) = $t\sqrt{25-t^{2}}$

f'(t) = $1.\sqrt{25-t^{2}}+\frac{t}{2}.(25-t^{2})^{-1/2}(-2t)$

f'(t) = $\sqrt{25-t^{2}} -\frac{t^{2}}{\sqrt{25-t^{2}} }$

f'(t) = $\frac{25-2t^{2}}{\sqrt{25-t^{2}} }$ = 0

For this function to be zero, only denominator must be zero:

$25-2t^{2} = 0$

t = ± $\sqrt{2.5}$

$\sqrt{25-t^{2}}$ ≠ 0

t = ± 5

Now, evaluate critical points in the given interval.

t = $-\sqrt{2.5}$ and t = - 5 don't exist in the given interval, so their f(x) don't count.

f(t) = $t\sqrt{25-t^{2}}$

f(-1) = $-1\sqrt{25-(-1)^{2}}$

f(-1) = $-\sqrt{24}$

f(-1) = $-2\sqrt{6}$

f( $\sqrt{12.5}$ ) = $\sqrt{12.5} \sqrt{25-(\sqrt{12.5} )^{2}}$

f( $\sqrt{12.5}$ ) = 12.5

f(5) = $5\sqrt{25-5^{2}}$

f(5) = 0

Therefore, absolute maximum is f( $\sqrt{12.5}$ ) = 12.5 and absolute minimum is

f(-1) = $-2\sqrt{6}$ .

8 0

3 years ago

An equation of a hyperbola is given.

siniylev [52]

Answer:

The vertices are $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

The foci are $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$ .

The asymptotes are $y=2x,\:y=-2x$ .

b) The length of the transverse axis is 6.

c) See below.

Step-by-step explanation:

$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$ is the standard equation for a right-left facing hyperbola with center $\left(h,\:k\right)$ .

The vertices $\:\left(h+a,\:k\right),\:\left(h-a,\:k\right)$ are the two bending points of the hyperbola with center $\:\left(h,\:k\right)$ and semi-axis a, b.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ and vertices $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

For a right-left facing hyperbola, the Foci (focus points) are defined as $\left(h+c,\:k\right),\:\left(h-c,\:k\right)$ where $c=\sqrt{a^2+b^2}$ is the distance from the center $\left(h,\:k\right)$ to a focus.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ $c=\sqrt{3^2+6^2}= 3\sqrt{5}$ and foci $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$

The asymptotes are the lines the hyperbola tends to at $\pm \infty$ . For right-left hyperbola the asymptotes are: $y=\pm \frac{b}{a}\left(x-h\right)+k$