Find the slope of a line through the points (-1,5) and (7,-11)

Mathematics

1 answer:

horrorfan [7]3 years ago

5 0

Answer:

-2

Step-by-step explanation:

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The Hudson Bay tides vary between 3 feet and 9 feet. The tide is at its lowest point when time (t) is 0 and completes a full cyc

BaLLatris [955]

Answer:

$y(t)= 6-3cos(\dfrac{2\pi}{14}t )$

Step-by-step explanation:

The function that could model this periodic phenomenon will be of the form

$y(t) = y_0+Acos(wt)$

The tide varies between 3ft and 9ft, which means its amplitude $A$ is

$A =\dfrac{(9-3)ft}{2} \\\\\boxed{A = 3ft}$

and its midline $y_0$ is

$y_o=3+3 \\\\\boxed{y_o= 6ft}$ .

Furthermore, since at $t=0$ the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is $-cos(\omega t)$ .

The period of the full cycle is 14 hours, which means

$\omega t =2\pi$

$\omega (t+14)= 4\pi$

giving us

$\boxed{\omega = \dfrac{2\pi}{14}.}$

With all of the values of the variables in place, the function modeling the situation now becomes

$\boxed{y(t)= 6-3cos(\dfrac{2\pi}{14}t ).}$

8 0

3 years ago

What is the relationship between BF and CD?https://www.pearsonrealize.com/community/proxy/assessment/38d9dab82a32446fbb58f979cee

Sunny_sXe [5.5K]

These 2 lines never touch each other

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3 years ago

Evaluate cos(sin^-1(4/5)

PSYCHO15rus [73]

Step-by-step explanation:

$Let \: \: \theta = \sin^{ - 1} (\frac{4}{5}) \\ \\ \therefore \: \sin\theta = \frac{4}{5} \\ \\ \because \: { \cos}^{2} \theta =1 - { \sin}^{2} \theta \\ \\ \therefore \: { \cos}^{2} \theta = 1 - (\frac{4}{5} )^{2} \\ \\ = 1 - \frac{16}{25} \\ \\ = \frac{25 - 16}{25} \\ \\ = \frac{9}{25} \\ \\ \therefore \:{ \cos}\theta = \pm \: \frac{3}{5} \\ \\ \because \: \theta \: lie \: in \: the \: first \: quadrant \\ \\ \therefore \: { \cos}\theta = \: \frac{3}{5} \\ \\ \implies \theta = { \cos}^{ -1 } \: \frac{3}{5}\\\\\implies \theta = { \cos}^{ -1 } \: \frac{3}{5}= { \sin}^{ -1 } \: \frac{4}{5} \\ \\ \therefore \: \cos( \ {sin}^{ - 1} \frac{4}{5} ) \\ \\ = \cos( \ {cos}^{ - 1} \frac{3}{5} ) \\\\ = \frac{3}{5} \\ \\ \purple{ \boxed{\therefore \: \cos( \ {sin}^{ - 1} \frac{4}{5} ) = \frac{3}{5}}}$