10.125 as percent help please

Mathematics

2 answers:

devlian [24]3 years ago

7 0

10.125 as a percentage is 1012.5%

vodomira [7]3 years ago

6 0

Answer:

10.125 = 1012.5%

Step-by-step explanation:

$p\%=\dfrac{p}{100}$

therefore

$a=a\cdot100\%$

----------------------------------------------

$10.125=10.125\cdot100\%=1012.5\%$

You might be interested in

HELP ME PLEASE!!! I SWEAR I'LL GIVE YOU THE BRAINLIEST ANSWER

Pepsi [2]

Answer:

118

Step-by-step explanation:

Total surface area=2(lb+bh+lb)

=2(3.5×4+4×6+3.5×6)

=2(14+24+21)

=2×59

=118

7 0

4 years ago

The scatter plot shows the relationship between the number of car accidents in a month and the number of drivers attending a pro

Dovator [93]

For the answer to the question above, the relationship between the number of car accidents in a month and the number of drivers attending a program on distracted driving:y=−0.67x+17
-0.67 is the slope, it is negative so it is decreasing
this represents The number of accidents was reduced by 0.67 per month every month.

8 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP(25 points)

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:

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

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

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

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

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