Identity the slope and y-intercept<br> (-2,-3) (0,5)

A phone company worker needs to know the height of a cell tower, so he decides to use a 2 m pole and shadows cast by the sun, as

olganol [36]

I always found it easiest to draw out the picture

6 0

3 years ago

Find the exact value of cos(sin^-1(-5/13))

son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

$\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}$

$\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}$

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

8 0

2 years ago

A student receives test scores of 62, 83, and 91. This student's term project score is 88. Each test is worth 20% of the final g

miskamm [114]

Answer:

The student's current average score will be 69.2

Step-by-step explanation:

Let first test be TEST A: which is 20 of total and secures 62

Let second test be TEST B: which is 20 of the total marks and has secured 83.

Let third test be TEST C: which is 20 of total and has secured 91.

And now the TEST D which is 25 of total and has secured 88.

Therefore, by multipying across