A line passes through the points (0,5) and (3,6). What is its equation in slope-intercept

Mathematics

1 answer:

alexandr402 [8]3 years ago

8 0

Answer:

y = 1/3x + 5

Step-by-step explanation:

Slope = (change in y)(change in x)

Slope = (6 - 5)/(3 - 0)

Slope = 1/3

Substitute m = 1/3 and one given ordered pair into y = mx + b. using (3, 6)

y = mx+ b

6 = 1/3(3) + b

6 = 1 + b

5 = b

the equation is y = 1/3x + 5

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miskamm [114]

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:

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

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

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

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

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