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Veronika [31]
3 years ago
12

Select the graph that represents two quantities in a proportional relationship.

Mathematics
1 answer:
julia-pushkina [17]3 years ago
6 0

Answer:

B

Step-by-step explanation:

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Anna=12x because Demarco is 4x and Anna is 3 times older than 4x
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Help Plis.......................
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Step-by-step explanation:

Two negative signs make a positive number. So -(-4) makes 4. I know it sounds weird, but thats what schools teach. Hope this helps!

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A) 9x^2-1<br> B)9x2-6x+1<br> C)3x^2-1<br> D) 3x-1
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3 years ago
HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP(25 points)
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Answer: I hope it helps :)

  • x=6 , y=6√3
  • x =23√3 , y=23
  • u =12 , v= 6
  • a =18√2 , b =18
  • x = 13 , y= 13

Step-by-step explanation:

1.

Hypotenuse =x\\Opposite =y \\Adjacent =6\\\alpha = 6\\Let's\: find\: the \:hypotenuse\: first\\Using SOHCAHTOA\\Cos \alpha = \frac{adj}{hyp} \\Cos 60 = \frac{6}{x} \\\frac{1}{2} =\frac{6}{x} \\Cross\:Multiply\\x = 12\\Let's\: find\: y\\Hyp^2=opp^2+adj^2\\12^2=y^2+6^2\\144=y^2+36\\144-36=y^2\\108=y^2\\\sqrt{108} =\sqrt{y^2} \\y=6\sqrt{3}

2.

Opposite =x\\Hypotenuse = 46\\Adjacent =y \\\alpha =60\\Using \: SOHCAHTOA\\Sin \alpha =\frac{opp}{adj} \\Sin 60=\frac{x}{46}\\\\\frac{\sqrt{3} }{2} =\frac{x}{46}  \\2x=46\sqrt{3} \\x = \frac{46\sqrt{3} }{2} \\x =23\sqrt{3} \\\\Hyp^2=opp^2+adj^2\\46^2=(23\sqrt{3} )^2+y^2\\2116=1587+y^2\\2116-1587=y^2\\529=y^2\\\sqrt{529} =\sqrt{y^2} \\y = 23

3.

Hypotenuse = u\\Opposite =6\sqrt{3} \\Adjacent = v\\\alpha =60\\Sin\: 60 = \frac{6\sqrt{3} }{u} \\\frac{\sqrt{3} }{2} =\frac{6\sqrt{3} }{u} \\12\sqrt{3} =u\sqrt{3} \\\\\frac{12\sqrt{3} }{\sqrt{3} } =\frac{u\sqrt{3} }{\sqrt{3} } \\u = 12\\Hyp^2=opp^2+adj^2\\12^2= (6\sqrt{3} )^2+v^2\\144=108+v^2\\144-108=v^2\\36 = v^2\\\sqrt{36} =\sqrt{v^2} \\\\v =6

4.

Hypotenuse = a\\Opposite =18 \\Adjacent = b\\\alpha =45\\Tan \alpha = opp/adj\\Tan \:45 =18/b\\1=\frac{18}{b}\\ b = 18\\\\Hyp^2=Opp^2+Adj^2\\a^2 = 18^2+18^2\\a^2=324+324\\a^2=648\\\sqrt{hyp^2} =\sqrt{648}\\ \\a =18\sqrt{2}

5.

Hypotenuse = 13\sqrt{2}\\ Opposite =x\\Adjacent = y\\\alpha =45\\Sin\:\alpha = opp/hyp\\Sin 45=x/13\sqrt{2}\\ \\\frac{\sqrt{2} }{2} =\frac{x}{13\sqrt{2} } \\2x=26\\2x/2=26/2\\\\x = 13\\\\Hyp^2=opp^2+adj^2\\(13\sqrt{2})^2=13^2+y^2\\ 338=169+y^2\\338-169=y^2\\169=y^2\\\sqrt{169} =\sqrt{y^2} \\13 = y

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A and C would be the correct answer
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