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hoa [83]
1 year ago
14

Which equation represents the line that passes through the point (-2.4, -7.1) and has a slope of -5.3

Mathematics
1 answer:
Wewaii [24]1 year ago
8 0
Y= -5.3x + c

-7.1= -5.3(-2.4) + c

C= -12.72-7.1 = −19.82

Y= —5.3x-19.82
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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:

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

7 0
3 years ago
Find these absolute values 8
Amiraneli [1.4K]

Answer:

the answer is |-8| and |8|

Step-by-step explanation:

both of the answers in absolute value will give 8

5 0
3 years ago
Use​ Green's Theorem to evaluate the line integral. Assume the curve is oriented counterclockwise. Contour integral Subscript Up
Setler79 [48]

By Green's theorem, we have

\displaystyle\int_C\left(7x+\cos\frac1y\right)\,\mathrm dy-(3y^2+\ln(3x))\,\mathrm dx

=\displaystyle\iint_{[1,4]\times[0,3]}\frac{\partial\left(7x+\cos\frac1y\right)}{\partial x}-\frac{\partial(-(3y^2+\ln(3x))}{\partial y}\,\mathrm dx\,\mathrm dy

=\displaystyle\int_0^3\int_1^47+6y\,\mathrm dx\,\mathrm dy=\boxed{144}

4 0
3 years ago
The acceleration function a(t) (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.
klemol [59]

Answer:

A: v(t)=-t^2+4t-5

Step-by-step explanation:

Acceleration is second derivative of position, velocity is first derivative. Therefore, the velocity is the integral of acceleration.

v(t)=\int\ {2t+4} \, dt

Integrate:

-t^2+4t+C

V(0)=-5:

-0^2+4(0)+C=-5\\C=-5

Therefore, v(t):

v(t)=-t^2+4t-5

7 0
3 years ago
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lesantik [10]

Answer:

∠ 1 = 20 x + 5

∠ 2 = 24 x - 1

If lines l and m are parallel:

∠ 1 + ∠ 2 = 180°

20 x + 5 + 24 x - 1 = 180

44 x + 4 = 180

44 x = 180 - 4

44 x = 176

x = 176 : 44

x = 4°

Your welcome again :)

Step-by-step explanation:

6 0
3 years ago
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