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Katena32 [7]
3 years ago
11

Why is adding 3/9 and 6/9 different from multpying the two fractions? Explain

Mathematics
1 answer:
tankabanditka [31]3 years ago
4 0
<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ When you add fractions, you have to have the same denominator whereas, when you multiply them, you don't have to. (In this case the denominators are already the same.) Also, multiplying the fractions would make the result smaller whereas adding them would make it larger.

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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Area of a triangle with points at (-9,5), (6,10), and (2,-10)
Ann [662]
First we are going to draw the triangle using the given coordinates. 
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Distance from point A to point B:
d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}
d_{AB}= \sqrt{(6+9)^2+(10-5)^2}
d_{AB}= \sqrt{(15)^2+(5)^2}
d_{AB}= \sqrt{225+25}
d_{AB}= \sqrt{250}
d_{AB}=15.81

Distance from point A to point C:
d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}
d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}
d_{AC}= \sqrt{11^2+(-15)^2}
d_{AC}= \sqrt{121+225}
d_{AC}= \sqrt{346}
d_{AC}= 18.60

Distance from point B from point C
d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}
d_{BC}= \sqrt{(-4)^2+(-20)^2}
d_{BC}= \sqrt{16+400}
d_{BC}= \sqrt{416}
d_{BC}=20.40

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
s= \frac{AB+AC+BC}{2}
s= \frac{15.81+18.60+20.40}{2}
s= \frac{54.81}{2}
s=27.41

Finally, to find the area of our triangle, we are going to use Heron's formula:
A= \sqrt{s(s-AB)(s-AC)(s-BC)}
A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}
A= \sqrt{27.41(11.6)(8.81)(7.01)}
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We can conclude that the perimeter of our triangle is 140.13 square units.

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2 years ago
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Step-by-step explanation:

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= (3×64)+(5×70)

----‐------------------

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=542/448

= 271/224

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Step-by-step explanation:

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