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faust18 [17]
4 years ago
12

Solve for x: x^19 − 2x^18 + 7x^17 − 12x^6 + 6x^15 = 0

Mathematics
1 answer:
Flauer [41]4 years ago
7 0
It looks like you intend
  x¹⁹ -2x¹⁸ +7x¹⁷ -12x¹⁶ +6x¹⁵ = 0
  x¹⁵(x -1)²(x² +6) = 0

x = 0, 1, ±(√6)i . . . . . . . selection D is appropriate
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Kisachek [45]

If two angles and included side of one triangle are equal to two angles and the included side of the other triangle , then the two triangles are congruent (ASA rule).

in figure

  • consider triangle ABD and ACD
  • Angle BAD = angle CAD
  • AB = AC
  • Angle B = angle C
  • triangle ABD congruent to Triangle ACD ( by ASA rule)
7 0
3 years ago
How many degrees does the polynomial 6c9 have
spayn [35]

Answer:

None, there are no degrees/exponents

Step-by-step explanation:

Hope this is helpful :)

6 0
3 years ago
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((MATHS FORM2 CHAPTER 5 CIRCLE)). HELPPPPPPPPPP I WILL MARK AS BRAINLIEST PLEASE SHOW YOUR WORKING :-)
katrin2010 [14]

Answer:

Step-by-step explanation:

strategy:  Find the area of the entire dartboard, then subtracted the inner circle from that whole.    we will have the entire outer circle,  so divide that by 2 for half of it.   Next divide the inner circle by 4 . Then add that to the outer circle answer from above.

whole circle = \pi*20^{2} = 1257.142857 cm^{2}

inner circle = \pi*10^{2} = 314.2857143 cm^{2}

1257.142857 - 314.2857143 = 942.8571427 outer circle

942.8571427 / 2 = 471.4285 ( half of the outer circle)

now find quarter of the inner circle

314.2857143 / 4 = 78.57142855

add the two parts together

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6 0
3 years ago
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Nezavi [6.7K]

Answer:

105 if the triangle is 180

Step-by-step explanation:

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3 years ago
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Determine the gravitational force of attraction between two 3.5 kg bowling balls whose
Ipatiy [6.2K]

Answer:

F_G=2.04\ .\ 10^{-10}\ Nw

Option J

Step-by-step explanation:

Newton's Universal Law Of Gravitation

Newton discovered that all objects attract each other with a force of gravitational attraction which is proportional to their masses and inversely proportional to the square of the distance between them.

\displaystyle F_G=G\frac{m_1m_2}{d^2}\ ,\ G=6.673\ .\ 10^{-11} N m^2/kg^2

Where m_1, m_2 are the masses in kg and d is the distance between them in meters.

In the problem, we have

m_1=m_2=3.5\ kg

d=2\ meters

\displaystyle F_G=6.673\ .\ 10^{-11} N m^2/kg^2\frac{(3.5kg)(3.5kg)}{(2m)^2}

\boxed{F_G=2.04\ .\ 10^{-10}\ Nw}

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4 years ago
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