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irakobra [83]
3 years ago
8

Find m angle 1 and m angle 3 in the kite.

Mathematics
1 answer:
stepladder [879]3 years ago
6 0

Answer:

m∠1=39°, m∠3=51°

Step-by-step explanation:

Angle 3 = 180-(39+90)

180-129=51

Angle 1 = 39 b/c of the corresponding angles.

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What is the equation of the line that passes through (6,4) and (4,1)
Murrr4er [49]

Answer:

y = 3/2*x - 5

Step-by-step explanation:

Lets say that P1=(6,4) and P2=(4,1) and that the form of the equation must be y=m*x+b where m is the slope and b the independent variable. Then having two given points we can use the slope formula to find the slope value as:

P1=(x1,y1) and P2=(x2,y2)

slope formula --->   m=(y2-y1)/(x2-x1)

Replacing the given points --->    m=(1-4)/(4-6) = 3/2

then replacing the slope value obtained:

y = 3/2*x + b

Now lets find the value of b. For this we have to replace in the equation a point it can be P1 or P2, i will replace P2:

1 = 3/2*4 + b

1 = 6 + b

1 - 6 = b

-5 = b

therefore the line equation is:

y = 3/2*x - 5

3 0
3 years ago
How do you find a vector that is orthogonal to 5i + 12j ?
Rashid [163]
\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies  -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-------------------------------\\\\

\bf \boxed{5i+12j}\implies 
\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}
\\\\\\
slope=\cfrac{12}{{{ 5}}}\qquad negative\implies  -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}
\\\\\\
\ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}&#10;\\\\\\&#10;\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}&#10;\\\\\\&#10;\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}
4 0
4 years ago
if the hypotenuse of a special triangle of 45 45 90 was 10 radical 6 how would u solve for the hypotenuse
Vinvika [58]
Just do it it’s quite easy
3 0
4 years ago
You are building a rectangular dog pen with an area of 90ft^2. You want the length to be 3 ft longer that twice it's width. What
Alecsey [184]
I think the answers is the you think is if you think okayω
8 0
3 years ago
HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! PLZZZZZZZZZ 8PTS!!!!!!!!!!!!!!!!
Harlamova29_29 [7]

Answer:

The first box is c

second box is b

Third box is c

Step-by-step explanation:

6 0
4 years ago
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