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raketka [301]
3 years ago
7

The interior angles of a triangle have measures k°, 27°, and 10°. What is the value of k?ANSWER MY FREAKING QUESTION

Mathematics
1 answer:
lapo4ka [179]3 years ago
6 0
143o

interior angles of a triangle always add up to 180o
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A point is randomly chosen on a map of North America. Describe the probability of the point being in each location: North Americ
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Answer:

We know that the map is of North America:

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or 100% in percentage form.

2) New York City.

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3) Europe:

As this is a map of Noth America, you can not randomly point at Europe in it (Europe is other continent).

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5 0
3 years ago
A. x-int:-0.5,y-int:1<br>B. x-int:0.5,y-int:1<br>C. x-int:-0.5,y-int:-1<br>D. x-int:1,y-int:0.5
hodyreva [135]
The correct answer is:  [D]:  " <span>x-int : 1 ,  y-int:  0.5  " .
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In other words, what is (are) the point(s) of the graph at which "x = 0<span>" ?
</span>
By examining the graph, we see that when " x = 0" ; y is equal to:  "1<span>" .
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So; the "x-intercept" is at point:  "(0, 1)" ; or, we can simply say that the 
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_________________________________________________________</span>  Note:
_____________________________________________________
The "y-intercept" refers to the point(s) at which the the graph of a function (which is a line, in this case) cross(es) the "x-axis".  

In other words, what is (are) the point(s) of the graph at which " y = 0 <span>" ?
</span>
By examining the graph, we see that when " y = 0 " ; x  is equal to:  "0.5<span>" .
</span>
So; the "x-intercept" is at point:  "(0.5, 0)" ; or, we can simply say that the 
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______</span>_________________________________________________
This would correspond to:<span>
_______________________________________________________
           Answer choice:  [D]:  </span>" x-int: 1 , y-int: 0.5  " .
_______________________________________________________
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3 years ago
Find the polynomial f(x) of degree 3 with real coefficients that has a y-intercept of 60 and zeros 3 and 1+3i.
Sauron [17]

\bf \begin{cases} x=3\implies &x-3=0\\ x=1+3i\implies &x-1-3i=0\\ x=1-3i\implies &x-1+3i=0 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ (x-3)(x-1-3i)(x-1+3i)=0 \\\\\\ (x-3)\underset{\textit{difference of squares}}{([x-1]-3i)([x-1]+3i)}=0\implies (x-3)([x-1]^2-[3i]^2)=0 \\\\\\ (x-3)([x^2-2x+1]-[3^2i^2])=0\implies (x-3)([x^2-2x+1]-[9(-1)])=0

[ correction added, Thanks to @stef68 ]

\bf (x-3)([x^2-2x+1]+9)=0\implies (x-3)(x^2-2x+10)=0 \\\\\\ x^3-2x^2+10x-3x^2+6x-30=0\implies x^3-5x^2+16x-30=f(x) \\\\\\ \stackrel{\textit{applying a translation with a -2f(x)}}{-2(x^3-5x^2+16x-30)=f(x)}\implies -2x^3+10x^2-32x+60=f(x)

5 0
3 years ago
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