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2 years ago

12

Kaylee tell Aimee that the shape is a stop sign has a least one right angle. Aimee says that three are no right angles.who is ri

ght? explain

1 answer:

damaskus [11]2 years ago

4 0

Aimee is correct because a stop sign is an octogon. Octogons contain only obtuse angles.

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Find the Y-coordinate of point P that lies 1/3 along segment RS, where R (-7, -2) and S (2, 4).

QveST [7]

Solution:

Given that the point P lies 1/3 along the segment RS as shown below:

To find the y coordinate of the point P, since the point P lies on 1/3 along the segment RS, we have

$\begin{gathered} RP:PS \\ \Rightarrow\frac{1}{3}:\frac{2}{3} \\ thus,\text{ we have} \\ 1:2 \end{gathered}$

Using the section formula expressed as

$[\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}]$

In this case,

$\begin{gathered} m=1 \\ n=2 \end{gathered}$

where

$\begin{gathered} x_1=-7 \\ y_1=-2 \\ x_2=2 \\ y_2=4 \end{gathered}$

Thus, by substitution, we have

$\begin{gathered} [\frac{1(2)+2(-7)}{1+2},\frac{1(4)+2(-2)}{1+2}] \\ \Rightarrow[\frac{2-14}{3},\frac{4-4}{3}] \\ =[-4,\text{ 0\rbrack} \end{gathered}$

Hence, the y-coordinate of the point P is

$0$

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Solve 2cos ²y -siny -1=0 for 0° ≤y≤360°

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$\displaystyle\bf\\\boxed{\bf sin\,y=x}\\\\2x^2+x-1=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4\cdot2\cdot(-1)}}{2\cdot2}=\\\\=\frac{-1\pm\sqrt{1+8}}{4}=\frac{-1\pm\sqrt{9}}{4}=\frac{-1\pm3}{4}\\\\x_1=\frac{-1+3}{4}=\frac{2}{4}=\boxed{\bf\frac{1}{2}}\\\\x_2=\frac{-1-3}{4}=\frac{-4}{4}=\boxed{\bf-1}\\\\sin\,y=\frac{1}{2}\\\\\boxed{\bf y_1=\frac{\pi}{6}~~or~~(30^o)}\\\\\boxed{\bf y_2=\frac{5\pi}{6}~~or~~(150^o)}\\\\sin\,y=-1\\\\\boxed{\bf y_3=\frac{3\pi}{2}~~or~~(270^o)}$

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