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Ostrovityanka [42]
3 years ago
11

How do you find the area of a triangle?

Mathematics
2 answers:
Doss [256]3 years ago
8 0
Its base × hight / 2
guapka [62]3 years ago
7 0
The formula to find the area of a triangle is

\frac{h_{b}b }{2}

If you have any more questions let me know!
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Maddy and her cousin work during the summer for a landscaping company. Maddy's cousin has been working for the company longer, s
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Hope it helps u mate. Plz follow me :)

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3 years ago
Subtract (x² + 5x – 6) – (3x^2 – 7x + 1).<br> What is the difference?
Lorico [155]
1. Distribute the Negative Signs
2. Combine like terms
-2x^2 + 12x-7
4 0
3 years ago
Solve the following equation for B be sure to take into account whether a letter is capitalized or not.
const2013 [10]

Answer:

B = <u>d - 3f²</u>

         5M

Step-by-step explanation:

d -3f² = 5 B B

5 BM = d - 3f²

B = <u>d - 3f²</u>

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6 0
3 years ago
Read 2 more answers
In a test of a​ gender-selection technique, results consisted of 229 baby girls and 7 baby boys. Based on this​ result, what is
kirill115 [55]

Answer:

I'm pretty sure that the answer would be 229/236.

Step-by-step explanation:

Ok so. Since the results of the test is 229 for a girl and the results for a boy is 7. So i'm pretty sure what to do is combine both the results and keep that as the denominator. And then the numerator will be the result. So in this case the numerator would be 229 and the denominator would be 236.

I hope this helped, have a great day!!!

3 0
3 years ago
I need help I can’t figure it out
dusya [7]

Answer:

\large\boxed{\dfrac{41p}{70q^4(r-9)^2}}

Step-by-step explanation:

\dfrac{2p^4}{5q^5(r-9)^3}\cdot\dfrac{41q(r-9)}{28p^3}\\\\=\dfrac{2\!\!\!\!\diagup^1\cdot41}{5\cdot28\!\!\!\!\!\diagup_{14}}\cdot\dfrac{q\!\!\!\!\diagup}{q^{5\!\!\!\!\diagup^4}}\cdot\dfrac{(r\!\!\!\!\!\!{--}-9)\!\!\!\!\!\!{--}}{(r-9)^{3\!\!\!\!\diagup^2}}\cdot\dfrac{p^{4\!\!\!\!\diagup}}{p^3\!\!\!\!\!\!\diagup}\\\\=\dfrac{41}{70}\cdot\dfrac{1}{q^4}\cdot\dfrac{1}{(r-9)^2}\cdot\dfrac{p}{1}\\\\=\dfrac{41p}{70q^4(r-9)^2}

7 0
3 years ago
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