PLEASE HELP

1 answer:

3 0

M = -5/3

slope formula:
m = y2 - y1 / x2 - x1

A) m = 10-13 / 17-12 = -3/5
B) m = 10-15 / 13-16 = -5/-3
C) m = 10-7 / 3 -0 = 3/3 = 1
D) m = 18-13/8-11 = 5/-3 or -5/3

D. (11,13); (8,18) are the two points the line could pass

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Answer the above question

Svetach [21]

Answer:

Using Pyhagoras Theorem for the right-angled triangle ABO,

h² = 12² + 16² = 400.

∴ The length of the side of the rhombus is 20 cm.

★½×24×32 = 20×h

∴ h= 19.2 cm.

3 0

3 years ago

Read 2 more answers

The angle θ lies in Quadrant II .

Andreyy89

let's keep in mind that, in the II Quadrant, cosine is negative and sine, is positive.

cosine is adjacent/hypotenuse, however the hypotenuse is simply a radius unit, and thus is never negative, so in the -(2/3) the negative must be the numerator, -2.

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-2}}{\stackrel{hypotenuse}{3}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{3^2-(-2)^2}=b\implies \pm\sqrt{5}=b\implies \stackrel{\textit{II Quadrant}}{+\sqrt{5}=b}~\hfill tan(\theta )=\cfrac{\stackrel{opposite}{\sqrt{5}}}{\stackrel{adjacent}{-2}} \\\\\\ ~\hspace{34em}$