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QveST [7]
2 years ago
14

7^1 in expanded form​

Mathematics
1 answer:
In-s [12.5K]2 years ago
8 0

Step-by-step explanation:

{7}^{1}  = 7 \times 1 \\  = 7

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LenKa [72]

Answer:

the one you have marker is correct

Step-by-step explanation:

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2 years ago
Which expression is equivalent to 9(-4b - 3)?<br> -4b - 27<br> -36b - 27<br> -27b - 36<br> -36b - 3
jenyasd209 [6]
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Sindrei [870]

Answer:

x=16\dfrac{28}{37}\\ \\y=5\dfrac{1}{3}\\ \\z=7.5

Step-by-step explanation:

The diagram shows three paralel lines which divide the large triangle into three similar triangles.

By Thales theorem,

\dfrac{5}{z}=\dfrac{y}{8}=\dfrac{2}{3}

Then

\dfrac{5}{z}=\dfrac{2}{3}\Rightarrow 2z=15,\ z=7.5\\ \\\dfrac{y}{8}=\dfrac{2}{3}\Rightarrow 3y=16,\ y=\dfrac{16}{3}=5\dfrac{1}{3}

From the similarity of triangles,

\dfrac{x}{20}=\dfrac{z+8}{z+8+3}\\ \\\dfrac{x}{20}=\dfrac{7.5+8}{7.5+8+3}\\ \\\dfrac{x}{20}=\dfrac{15.5}{18.5}\\ \\x=20\cdot \dfrac{155}{185}=20\cdot \dfrac{31}{37}=\dfrac{620}{37}=16\dfrac{28}{37}

6 0
3 years ago
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4vir4ik [10]

Answer:

  • 1/xy

Step-by-step explanation:

<u>Simplify the numerator:</u>

  • (2x - y)/(2x + y) + (2x + y)/(3y - 6x) + 8xy/(12x² - 3y²) =
  • (2x - y)/(2x + y) - (2x + y)/3(2x - y) + 8xy/3(2x + y)(2x - y) =
  • [3(2x - y)² - (2x + y)² + 8xy] / [3(2x + y)(2x - y)] =
  • [12x² - 12xy + 3y² - 4x² - 4xy - y² + 8xy] / [3(2x + y)(2x - y)] =
  • [8x² - 8xy + 2y²] / [3(2x + y)(2x - y)] =
  • 2{2x - y)² / [3(2x + y)(2x - y)] =
  • 2(2x - y) / [3(2x + y)]

<u>Simplify the denominator:</u>

  • (4x²y - 2xy²) / (6x + 3y) =
  • 2xy(2x - y) / [3(2x + y)]

<u>Now simplify the remainder of the expression:</u>

  • 2(2x - y) / [3(2x + y)] × [3(2x + y)]/[2xy(2x - y}] =
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4 0
2 years ago
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Pie

Answer:

Approximate answer: 66.36 \ cm^2

Step-by-step explanation:

DM = 5.66\cdot \cot(65^\circ)

Denote the projection of point B onto side DC as M'.

M'C = 5.66\cot(62^\circ)

MM' = AB = 8.9

DC = DM + MM' + M'C.

Next we should put everything together in area formula for trapizium:

S = AM\cdot\frac{AB + DC}{2}

4 0
3 years ago
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