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Sauron [17]
3 years ago
9

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

Mathematics
1 answer:
Naddik [55]3 years ago
4 0

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.

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Please answer this and will marked as brainlist ​
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<h3>Answer: 1</h3>

where x is nonzero

=======================================================

Explanation:

We'll use two rules here

  • (a^b)^c = a^(b*c) ... multiply exponents
  • a^b*a^c = a^(b+c) ... add exponents

------------------------------

The portion [ x^(a-b) ]^(a+b) would turn into x^[ (a-b)(a+b) ] after using the first rule shown above. That turns into x^(a^2 - b^2) after using the difference of squares rule.

Similarly, the second portion turns into x^(b^2-c^2) and the third part becomes x^(c^2-a^2)

-------------------------------

After applying rule 1 to each of the three pieces, we will have 3 bases of x with the exponents of (a^2-b^2),  (b^2-c^2) and (c^2-a^2)

Add up those exponents (using rule 2 above) and we get

(a^2-b^2)+(b^2-c^2)+(c^2-a^2)

a^2-b^2+b^2-c^2+c^2-a^2

(a^2-a^2) + (-b^2+b^2) + (-c^2+c^2)

0a^2 + 0b^2 + 0c^2

0+0+0

0

All three exponents add to 0. As long as x is nonzero, then x^0 = 1

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Step-by-step explanation:

Let n represent the number we're looking for. Then ...

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  n = (3 × 4.9 mm)/(0.7 mm) = 3 × 49/7 = 3 × 7

  n = 21

You would have to line up 21 amoeba proteus to equal the length of three pelomyxa palustris.

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