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LenaWriter [7]
1 year ago
10

Rewrite 7^5/7 with a single positive exponent

Mathematics
1 answer:
vaieri [72.5K]1 year ago
8 0

Answer:

7^4

Step-by-step explanation:

to divide exponents we just subtract what the exponent is so we can write it like this

7^5/7^1= 7^5-1=7^4

Hopes this helps please mark brainliest

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Answer:

  (2.27, 0.96)

Step-by-step explanation:

The orthocenter is the point of intersection of altitudes of the triangle. The equation for an altitude is the equation of a line through a vertex that is perpendicular to the opposite side.

For example, the line perpendicular to side AC can be found as ...

  (∆x, ∆y) = C-A = (3, -1) -(-4, 2) = (7, -3)

  ∆x(x -h) +∆y(y -k) = 0 . . . . perpendicular through point (h, k)

For side AC, we want the point to be B(4, 5), so the equation is ...

  7(x -4) -3(y -5) = 0

  7x -3y -13 = 0

Similarly, the altitude to side AB can be written as ...

  (∆x, ∆y) = B -A = (4, 5) -(-4, 2) = (8, 3)

  8(x -3) +3(y +1) = 0

  8x +3y -21 = 0

__

By the "cross multiplication method", the solution to these equations is ...

  x = (-3(-21) -(3(-13))/(7(3) -8(-3)) = (63+39)/(21+24) = 102/45 = 2 4/15 ≈ 2.27

  y = (-13(8) -(-21)(7))/45 = 43/45 ≈ 0.96

The orthocenter is near (2.27, 0.96).

__

A graphing application confirms this result.

_____

<em>Additional information about the cross multiplication method</em>

For the general form equations ...

  • ax +by +c = 0
  • dx +ey +g = 0

The "cross multiplication method" has you write the array ...

  \begin{array}{cccc}a&b&c&a\\d&e&g&d\end{array}

and form the "cross products" in groups of four coefficients:

  D = ae -db, X = bg -ec, Y = cd -ga

Then the solution to the set of equations is ...

  1/D = x/X = y/Y   ⇒   x = X/D, y = Y/D

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<em>Additional note</em>

Videos of this method show you writing the array as bcab/egde and using the final equation x/X=y/Y=1/D. The above gets the same result in a more straightforward manner.

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