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Leokris [45]
3 years ago
7

Find the value of x of the picture below HELP PLEASE ! -10 -15 -20 -25

Mathematics
2 answers:
garri49 [273]3 years ago
8 0
(180 - 5x + 12) + x + 3x + 8 = 180
180 - x + 20 = 180
-180
-x + 20 = 0
- 20
-x = -20
x = 20
Sorry for the confusion! I believe this is the correct answer
Vladimir79 [104]3 years ago
3 0
5x+12=3x+8+x(exterior angle property)
5x-4x=8-12
x=-4

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Brut [27]

Check the picture below.

let's recall that a kite is a quadrilateral, and thus is a polygon with 4 sides

sum of all interior angles in a polygon

180(n - 2)      n = number of sides

so for a quadrilateral that'd be 180( 4 - 2 ) = 360, thus

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LUCKY_DIMON [66]

Answer:

=8\sqrt{15}b^{\frac{7}{2}}

Step-by-step explanation:

\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}

=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}

=\sqrt{40}b^2\sqrt{24b^3}

\sqrt{24b^3}

=\sqrt{24}\sqrt{b^3}

=\sqrt{24}b^{\frac{3}{2}}

=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2

=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2

=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

\sqrt{2^3}

=2^{3\cdot \frac{1}{2}

=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}

2^{\frac{3}{2}+\frac{3}{2}}

=2^3

=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}

b^{\frac{3}{2}+2}

=b^{\frac{7}{2}}

=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}

=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}

=8\sqrt{15}b^{\frac{7}{2}}

4 0
3 years ago
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