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3 years ago

Please help I will give points

Mathematics

1 answer:

snow_lady [41]3 years ago

8 0

Answer:

1/40

Step-by-step explanation:

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How do i figure out these problems ?

Andreyy89

Given two vectors

$v=(a,b),\quad w=(c,d)$

they are parallel if they are a multiple of each other:

$c=ka,\quad d=kb$

You can easily test this by checking if

$\dfrac{c}{a}=\dfrac{d}{b}$

they are orthogonal if their dot product is null:

$v\cdot w=ac+bd=0$

For example, in the first case, we have

$\dfrac{10}{3}\div 2 \neq \dfrac{4}{3}\div 5$

So, they aren't parallel. Similarly, you have

$2\cdot \dfrac{10}{3}+5\cdot \dfrac{4}{3}\neq 0$

So, they aren't orthogonal.

In the second case, we have

$\dfrac{5}{-12}\neq \dfrac{-6}{-10}$

So, they aren't parallel, and

$5(-12)+(-6)(-10)=-60+60=0$

So, they are orthogonal.

Finally, we have

$\dfrac{2}{-4}=\dfrac{-7}{14}=-\dfrac{1}{2}$

So, they are parallel (and thus can't be orthogonal)

7 0

3 years ago

the table shows the ratio between the number of books ordered and their cost. 7,14 8,16 9,18. find equivalent ratios when the nu

Nadya [2.5K]

Answer:

Number of Books. Cost (dollars). 7 14. 8 16. 9 18. Find equivalent ratios when the number of books ordered is 1, 2, and 3. Write them as ordered pairs in the form...

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3 years ago

Hare $50 among Terry, Leroy and Johnson in the ratio 1:3:6. How much money does Johnson get? 1 point

nlexa [21]

Answer:

His share is $30

Step-by-step explanation:

Here, we are interested in knowing the amount of money that Johnson get.

From the ratio, we can see that Johnson’s ratio is the largest at 6

so to calculate his share of the $50, we add all the ratios together = 1 + 3 + 6 = 10

Now his share would be his ratio over the total * 50

= 6/10 * 50 = $30

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3 years ago

The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean μ, the act

cluponka [151]

Answer: 0.3061.

Step-by-step explanation:

Given : The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean $\mu$ , the actual temperature of the medium, and standard deviation $\sigma$ .