Please and Thank You

Mathematics

2 answers:

SCORPION-xisa [38]3 years ago

5 0

Answer:

It's B :D

Step-by-step explanation: I do not know--

VladimirAG [237]3 years ago

4 0

Answer:

The slope is 7. The y intercept is (0, -3)

Step-by-step explanation:

The slope is the number before the variable that determines how steep or shallow the line is. The y-intercept is determined by looking at the number after the variable which determines how high or low the line is from the origin.

You might be interested in

Graph the line with a slope of <img src="https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B3%7D" id="TexFormula1" title="\frac{5}{3}" alt

balu736 [363]

I don’t know sorry please somebody else can help you

8 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

3 years ago

Find the inverse Laplace transform f(t) of the function F(s). Write uc for the Heaviside function that turns on at c, not uc(t).

zzz [600]

Answer:

F(t) $=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}$