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3 years ago

PLEASE HELP ASAP!!!! IS THIS CORRECT?

Mathematics

2 answers:

xz_007 [3.2K]3 years ago

7 0

From what I am looking at, it seems as if it is. :)

stealth61 [152]3 years ago

6 0

No you got it wrong sorry and good luck :}

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If the length of each side of an equilateral triangle were increased by 50 percent, what would be the percent increase in the ar

Shkiper50 [21]

Answer:

The area increase by 225%

Step-by-step explanation:

attachment of a triangle

The area of a triangle equilateral is calculated with the next formula:

A= $\frac{a*h}{2}$

$h^{2} +(\frac{a}{2})^{2}$ = $a^{2}$

$h^{2} = a^{2} - \frac{a^{2} }{4}$ = $\frac{a^{2} }{4}$ *3

h= $\sqrt{\frac{a^{2}*3 }{4} }$

h= $\frac{\sqrt{3}*a }{2}$

replacing in the A equation:

A= $\frac{a*\sqrt{3}*a }{2*2}$ 

A= $\frac{\sqrt{3}*a^{2} }{4}$ 

Now each side increse by 50% ⇒ a=1.5a

A= $\frac{\sqrt{3}*(1.5a)^{2} }{4}$ 

A= $\frac{\sqrt{3}*a^{2}* 2.25 }{4}$ 

A=2.25 * $\frac{\sqrt{3}*a^{2} }{4}$ 

It means that the area incrase by 225%

6 0

3 years ago

How many atoms of group one would react with one atom of group five?

PSYCHO15rus [73]

Group 1 elements are comprised of: (1) hydrogen, sodium, lithium, potassium, rubidium, caesium, and francium. The tendency of these atoms is to lose one of their electrons such that they become charge +1.

Group 5 elements contains the vanadium, niobium, tantalum, and dubnium. It is quite hard to predict the oxidation number of these elements such that it will be hard to determine the number of atoms needed to react with them. Their valence electron is 5, and they react by losing them. Hence, in order to reach the octet rule, 3 group 1 elements should be reacted.

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3 years ago

The ordered pair (3,9) is located in what quadrant? Quadrant I Quadrant II Quadrant III Quadrant IV 2 points QUESTION 2 If you m

galben [10]

For the first question the answer would be the first quadrant

8 0

3 years ago

Read 2 more answers

6(5 + 3x) is equivalent too

Artyom0805 [142]

Answer:

30+18 x

Step-by-step explanation:

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2 years ago

Find the domain of f(x)=x-9/x squared-81

vagabundo [1.1K]

$f(x)=\dfrac{x-9}{x^2-81}\\\\\text{The domain:}\\\\x^2-81\neq0\qquad\text{add 81 to both sides}\\\\x^2\neq81\to x\neq\pm\sqrt{81}\\\\x\neq-9\ \wedge\ x\neq9\\\\Answer:\ \boxed{\text{All real numbers except -9 and 9}}\to\boxed{x\in\mathbb{R}-\{-9,\ 9\}}$

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3 years ago