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hram777 [196]
3 years ago
12

81^5=3^x What does “x” equal?

Mathematics
1 answer:
Lynna [10]3 years ago
7 0

Answer:

x =2 0

Step-by-step explanation:

{81}^{5}  =  {3}^{x}

We know that 3^4 = 81. So :

=  >  { ({3}^{4}) }^{5}  =  {3}^{x}

=  >  {3}^{4 \times 5}  =  {3}^{x}

=  >  {3}^{20}  =  {3}^{x}

Here both the Right hand side & Left hand side has same base i.e. have 3 as base in both the sides.

So , eliminating the base gives :-

x = 20

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If B is the midpoint of AC, solve for x, and find the lengths of AB, BC, and AC.
makvit [3.9K]

Answer:

49, 49 , 98

Step-by-step explanation:

AC = AB + BC = 2AB

  • 3x - 31 = 2(x+6)
  • 3x - 31 = 2x + 12
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AB= BC = x + 6 = 43 + 6 = 49

AC = 2AB = 2* 49 = 98

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3 years ago
The Earth is 1.49 × 108 kilometers from the Sun, multiply this expression by the given power of 10.
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7 0
3 years ago
Find the perimeter of pentagon STUVW WITH VERTICES S(0,0) T(3,-2) U(2,-5) V(-2,-5) W(-3,-2)
andrew11 [14]
Use the distance formula.

\sqrt{( x_{2} - x_{1} )^2 + (y_{2} - y_{1})^2}

 
Points S and W.
\sqrt{(3)^2 + (2)^2}

\sqrt{9+4}

\sqrt{13}

~3.6

Points S and T
\sqrt{(3 - 0)^2 + (-2 - 0)^2}

\sqrt{(3)^2 + (-2)^2}

\sqrt{9+4}

\sqrt{13}

~3.6

Points T and U
\sqrt{(3 - 2)^2 + (-2 + 5)^2}

\sqrt{(1)^2 + (3)^2}

\sqrt{1+9}

\sqrt{10}

~3.1

Points U and V
\sqrt{(2+2)^2 + (-5 + 5)^2}

\sqrt{(4)^2 + (0)^2}

\sqrt{16}

~4

Points V and W
\sqrt{(-2+3)^2 + (-5 + 2)^2}

\sqrt{(1)^2 + (-3)^2}

\sqrt{2+9}

\sqrt{11}

~3.3

Add all these together.

3.3 + 3.1 + 4 + 3.1 + 3.6
≈17
4 0
3 years ago
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