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3 years ago

Help with this please

Mathematics

1 answer:

Stella [2.4K]3 years ago

6 0

Answer:

1 and 7 and 5

Step-by-step explanation:

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Order from least to greatest: 7/12, 0.75, 5/6

Natalija [7]

From least the greatest the answer would be 7/12 .75 5/6

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3 years ago

Read 2 more answers

PORFAVOR NECESITO AYUDA CON LA PREGUNTAS ANTERIORES Q HICE PORFA AYUDENME!!!!

Black_prince [1.1K]

Answer:

Where are the questions?

Step-by-step explanation:

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3 years ago

18)

katovenus [111]

I'd say

Put one end of the compass on point P and the other end on point Q.

This way, you can construct (part of) a circumference with radius PQ with center P.

If you connect point P with any point on the circumference, this segment will be another radius of the circumference, and thus congruent to PQ.

7 0

3 years ago

A used car already has 40,000 miles on its odometer. The average owner will drive 12,000 miles per year. What function best mode

alina1380 [7]

The answer i got was 28000

7 0

3 years ago

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