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2 years ago

Find the value of x and give a reason for the relationship between the angles.

Mathematics

2 answers:

quester [9]2 years ago

8 0

Answer: ifk

Step-by-step explanation:

juin [17]2 years ago

3 0

Answer:

x=17

Step-by-step explanation:

We know that,

When two lines intersect, the pair of Vertically Opposite Angles are equal.

So,

$87=(5x+2)\\87-2=5x\\85=5x\\x=85/5\\x=17$

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What value of W is a solution to this equation? W + 7 = 12

Talja [164]

Answer:

w=5

Step-by-step explanation:

Due to the subtraction property of equality, you have to subtract 7 from both sides to isolate the W.

4 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:

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

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

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

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

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

7 0

3 years ago

Please help me with this :)<br>If fake answer, then will be reported!

arsen [322]

Answer:

10 miles less

Step-by-step explanation:

To get the direct distance first, we use the pythagorean theorem. Distance from Ferris to Butte directly would be the hypotenuse. By solving for the hypotenuse, we have 25 miles. Now the distance used previously is 35 mile. Hence, it (35 - 25 = 10 ) miles less.

3 0

2 years ago

What is the solution to he equation (y/y-4)-(4/y+4)=3^2/y^2-16

kicyunya [14]

Answer:

$y=\pm i \sqrt{7}$