PLEASE HELP WITH THIS ONE QUESTION

HELP PICS INCLUDED! WILL GIVE BRAINLIEST! SHOW WORK AND EXPLAIN

Naddik [55]

I assume you know about the dot product, and that for two vectors $\mathbf a$ and $\mathbf b$ , the angle between them $\theta$ satisfies

$\mathbf a\cdot\mathbf b=\|\mathbf a\|\|\mathbf b\|\cos\theta\iff\cos\theta=\dfrac{\mathbf a\cdot\mathbf b}{\|\mathbf a\|\|\mathbf b\|}$

Then the vectors are parallel if the angle between them is 0 or 180 degrees (0 or pi radians), which would make $\cos\theta=1$ or $\cos\theta=-1$ , respectively.

Part A)

$\vec v_1=\langle\sqrt3,1\rangle\implies\|\vec v_1\|=\sqrt{(\sqrt3)^2+1^2}=\sqrt4=2$

$\vec v_2=\langle-\sqrt3,-1\rangle=-\vec v_1\implies\|\vec v_2\|=\|\vec v_1\|=2$

$\vec v_1\cdot\vec v_2=(\sqrt3)(-\sqrt3)+(1)(-1)=-4$

Then the angle between $\vec v_1,\vec v_2$ is such that

$\cos\theta=\dfrac{-4}{(2)(2)}=-1\implies\theta=\pi\,\mathrm{rad}$

so these vectors are parallel ("antiparallel", more specifically, which means they are parallel but point in opposite directions).

Part B) involves the same computations:

$\vec u_1=\langle2,3\rangle\implies\|\vec u_1\|=\sqrt{2^2+3^2}=\sqrt{13}$

$\vec u_2$ has the same components but differing by sign and order, as $\vec u_1$ ; its magnitude remains the same, though:

$\vec u_2=\langle-3,-2\rangle\implies\|\vec u_2\|=\sqrt{(-3)^2+(-2)^2}=\sqrt{13}$

$\vec u_1\cdot\vec u_2=(2)(-3)+(3)(-2)=-12$

$\implies\cos\theta=\dfrac{-12}{(\sqrt{13})(\sqrt{13})}=-\dfrac{12}{13}\implies\theta=\cos^{-1}\left(-\dfrac{12}{13}\right)$

which is neither 0 nor pi, which means these vectors are not parallel.

4 0

4 years ago

Help me please ASAP

vredina [299]

Answer:

the correct answer is c

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

What is the domain of the function?

ivann1987 [24]

For a radical with an even root, like 2 in this case, if the radicand turns to a negative value, the result is just an imaginary value, which is another way to say, there's no such a root, no solution per se.

for that, let's first check when the radicand turns to 0 first.

$\bf -8x+8=0\implies -8x=-8\implies x=\cfrac{-8}{-8}\implies \boxed{x=1}$

so, that happens when x = 1, -8(1) +8, is just -8+8 or 0, ok.

so.. if "x" goes a bit higher, like say 2, -8(2) + 8 --> -16 + 8 --> -16

you'd get a negative value, and the radical doesn't have a root for that.

so... the domain, or values "x" can safely take without making the square root expression a negative value, are 1 or below, for example, if we use say -5

-8(-5) + 8 --> 40+8 --> 48 <------ a positive value

thus, (-∞, 1]

8 0

3 years ago

Solve for x.<br><br> - 6 ≥ 10 - 8 x

Burka [1]

Answer:

$x \geqslant 2 \: \: \: \: or \: \: \: \: 2 \leqslant x \\$