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3 years ago

15

What is the simplest form of 32/72

2 answers:

Anastasy [175]3 years ago

8 0

The answer would be 4/9.

Elena-2011 [213]3 years ago

5 0

$\displaystyle\\
 \frac{32}{72} = \frac{8\times 4}{8\times 9} = \boxed{\frac{4}{9} }$

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What is 23.4 x 0.4 equals

Hunter-Best [27]

9.36 is what it equals

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Distance= 2797.2 time= 0.15 speed

Nutka1998 [239]

Speed = Distance ÷ Time

Distance = 2797.2

Time = 0.15

2797.2 / 0.15 = 18,648

Speed = 18,648

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3 years ago

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If A(4 -6) B(3 -2) and C (5 2) are the vertices of a triangle ABC fine the length of the median AD from A to BC. Also verify tha

Gnoma [55]

Answer:

a) The median AD from A to BC has a length of 6.

b) Areas of triangles ABD and ACD are the same.

Step-by-step explanation:

a) A median is a line that begin in a vertix and end at a midpoint of a side opposite to vertix. As first step the location of the point is determined:

$D (x,y) = \left(\frac{x_{B}+x_{C}}{2},\frac{y_{B}+y_{C}}{2} \right)$

$D(x,y) = \left(\frac{3 + 5}{2},\frac{-2 + 2}{2} \right)$

$D(x,y) = (4,0)$

The length of the median AD is calculated by the Pythagorean Theorem:

$AD = \sqrt{(x_{D}-x_{A})^{2}+ (y_{D}-y_{A})^{2}}$

$AD = \sqrt{(4-4)^{2}+[0-(-6)]^{2}}$

$AD = 6$

The median AD from A to BC has a length of 6.

b) In order to compare both areas, all lengths must be found with the help of Pythagorean Theorem:

$AB = \sqrt{(x_{B}-x_{A})^{2}+ (y_{B}-y_{A})^{2}}$

$AB = \sqrt{(3-4)^{2}+[-2-(-6)]^{2}}$

$AB \approx 4.123$

$AC = \sqrt{(x_{C}-x_{A})^{2}+ (y_{C}-y_{A})^{2}}$

$AC = \sqrt{(5-4)^{2}+[2-(-6)]^{2}}$

$AC \approx 4.123$

$BC = \sqrt{(x_{C}-x_{B})^{2}+ (y_{C}-y_{B})^{2}}$

$BC = \sqrt{(5-3)^{2}+[2-(-2)]^{2}}$

$BC \approx 4.472$

$BD = CD = \frac{1}{2}\cdot BC$ (by the definition of median)

$BD = CD = \frac{1}{2} \cdot (4.472)$

$BD = CD = 2.236$

$AD = 6$

The area of any triangle can be calculated in terms of their side length. Now, equations to determine the areas of triangles ABD and ACD are described below:

$A_{ABD} = \sqrt{s_{ABD}\cdot (s_{ABD}-AB)\cdot (s_{ABD}-BD)\cdot (s_{ABD}-AD)}$ , where $s_{ABD} = \frac{AB+BD+AD}{2}$

$A_{ACD} = \sqrt{s_{ACD}\cdot (s_{ACD}-AC)\cdot (s_{ACD}-CD)\cdot (s_{ACD}-AD)}$ , where $s_{ACD} = \frac{AC+CD+AD}{2}$

Finally,

$s_{ABD} = \frac{4.123+2.236+6}{2}$

$s_{ABD} = 6.180$

$A_{ABD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}$

$A_{ABD} \approx 3.004$

$s_{ACD} = \frac{4.123+2.236+6}{2}$

$s_{ACD} = 6.180$

$A_{ACD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}$

$A_{ACD} \approx 3.004$

Therefore, areas of triangles ABD and ACD are the same.

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4 years ago

Use common factors to write two fractions equivalent to 6/42

Cerrena [4.2K]

19.) 14/18, 21/27

20.) 3/21, 2/14

21.) 6

22.)8/9

25.) 36/73

26.) 1/3=.33 2/8=.25 Eric is more successful

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3 years ago

Snow+snow+snow=30<br>Snow+stick+stick=20<br>Stick+2glass+2glass=9<br>Stick+1glass+snow=?

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Stick =5
1 glass =1
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