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Tanya [424]
2 years ago
9

How would I solve this i got 4.5

Mathematics
2 answers:
Aliun [14]2 years ago
6 0

Answer:

up \: is \: the \: angle \: bisector \\ so \: 54 = 5x - 6 \\ 5x = 54 + 6 \\ 5x = 60 \\ x =  \frac{60}{5}  = 12 \\ thank \: you

makkiz [27]2 years ago
5 0

Answer:

54 = 5x - 6 \\ 5x = 54 + 6 \\ 5x = 60 \\ x =  \frac{60}{5}  = 12.

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0.2588

Step-by-step explanation:

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When 8 is subtracted from a certain number and the result is multoplied by 3
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3 / m + 4 - 4 / m = 6 (i) Show that this equation can be written as 6m^2 + 25m+16= 0
Evgen [1.6K]

Answer:

Proved Below

Step-by-step explanation:

\sf \frac{3}{m+4} - \frac{4}{m} = 6\\LCM = m(m+4)\\Multiplying \ both \ sides \ by\ m(m+4)\\3m - 4(m+4) = 6m(m+4)\\3m-4m-16 = 6m^2+24m\\-m-16 = 6m^2+24m\\Adding\ m\ to\ both\ sides\\-16 = 6m^2+24m+m\\-16 = 6m^2+25m\\Adding \ 16 \ to \ both \ sides\\0 = 6m^2+25m+16\\OR\\6m^2+25m+16 = 0

Hence, Proved that \sf \frac{3}{m+4} - \frac{4}{m} = 6 is equivalent to \sf 6m^2+25m+16 = 0.

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The members of a high school band asked a number of students whether they would like blue, gold, or both for the uniforms for th
MaRussiya [10]

The values of a and b are a = 33% and b = 43% ⇒ last answer

Step-by-step explanation:

The venn-diagram contains:

  • A circle labeled blue 32
  • A circle labeled gold 25
  • The two circles are overlap with label 12 on the overlap

The table:

→ Band  :  blue  :  N.b  : total

→ gold   :  16%   :  a      :  49%

→ N.g     :  b       :  8%   :  51%

→ Total  :  59%  :  41%  :  100%

We need to find the values of a and b

From the table

16% represents the percentage of blue and gold uniforms (2nd row with 2nd column)

From venn-diagram

The common part of blue and gold label with 12 (overlap of two circles)

∴ 16% of the total number of students = 12

∵ 16% × x = 12

∴ \frac{16}{100} × x = 12

∴ 0.16 x = 12

- Divide both sides by 0.16

∴ x = 75

∵ x represents the total number of students

∴ The total number of students were asked is 75

From the table

b represents the percentage of blue but not gold uniforms (3rd row and 2nd column)

From venn-diagram

The part that has blue and not gold labeled with 32 (the part of the blue circle not in the gold circle)

∴ 32 of the total number of students like the blue but not gold

- Divide 32 by the total 75 , then multiply the quotient by 100%

   to find the percentage of blue but not gold

∴ b = \frac{32}{75} × 100% = 42.6666%

∴ b ≅ 43%

From the table

a represents the percentage of gold nut not blue uniforms (2nd row and 3rd column)

From venn-diagram

The part that has gold and not blue labeled with 25 (the part of the gold circle not in the blue circle)

∴ 25 of the total number of students like the gold but not blue

- Divide 25 by the total 75 , then multiply the quotient by 100%

   to find the percentage of gold but not blue

∴ a = \frac{25}{75} × 100% = 33.3333%

∴ a ≅ 33%

The values of a and b are a = 33% and b = 43%

To check your answer complete the table you will find the total of it is 100% ( for the columns 16% + 43% = 59% , 33% + 8% = 41% , then 59% + 41% = 100%, for the rows 16% + 33% = 49% , 43% + 8% = 51% , then 49% + 51% = 100%)

Learn more:

You can learn more about the probability in brainly.com/question/3756853

#LearnwithBrainly

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Your answer would be 3160.

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