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Elina [12.6K]
3 years ago
8

Suppose A and B represent two different school populations where A > B and A and B must be greater than 0. Which of the follo

wing expressions is the largest? Explain why. Show all work necessary.
2(A + B)
(A + B)2
A2 + B2
A2 − B2
Mathematics
1 answer:
Ulleksa [173]3 years ago
6 0
We can reject the last one: subtracting a non-zero value will result in a smaller value.

So now we have:

<span>2(A + B)
(A + B)2
A2 + B2

If all of them are mulptiplications, then they are all equivalent!

I mean by this, if what you meant is this:

</span>
<span>2*(A + B)
(A + B)*2
A*2 + B*2

If there is no sign, then the multiplication sign is implicit,


and all of these expressions say exactly the same: two of A and two of B.
</span>
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olga55 [171]

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3 years ago
Prove that a triangle with the sides a - 1 cm to root under a ​
vivado [14]

Question is Incomplete, Complete question is given below.

Prove that a triangle with the sides (a − 1) cm, 2√a cm and (a + 1) cm is a right angled triangle.

Answer:

∆ABC is right angled triangle with right angle at B.

Step-by-step explanation:

Given : Triangle having sides (a - 1) cm, 2√a and (a + 1) cm.  

We need to prove that triangle is the right angled triangle.

Let the triangle be denoted by Δ ABC with side as;

AB = (a - 1) cm

BC = (2√ a) cm

CA = (a + 1) cm  

Hence,

{AB}^2 = (a -1)^2 

Now We know that

(a- b)^2 = a^2+b^2 - 2ab

So;

{AB}^2= a^2 + 1^2 -2\times a \times1

{AB}^2 = a^2 + 1 -2a

Now;

{BC}^2 = (2\sqrt{a})^2= 4a

Also;

{CA}^2 = (a + 1)^2

Now We know that

(a+ b)^2 = a^2+b^2 + 2ab

{CA}^2= a^2 + 1^2 +2\times a \times1

{CA}^2 = a^2 + 1 +2a

{CA}^2 = AB^2 + BC^2

[By Pythagoras theorem]

a^2 + 1 +2a = a^2 + 1 - 2a + 4a\\\\a^2 + 1 +2a= a^2 + 1 +2a

Hence, {CA}^2 = AB^2 + BC^2

Now In right angled triangle the sum of square of two sides of triangle is equal to square of the third side.

This proves that ∆ABC is right angled triangle with right angle at B.

4 0
3 years ago
THIS IS SO IMPORTANT PLZ HELP
Varvara68 [4.7K]

Step-by-step explanation:

{b}^{2}  - 12a

where

a =  \frac{1}{4}  \\  \\ b = 8

{8}^{2}  - 12 \times  \frac{1}{4}  \\  \\ 64 - 3  \times 1 = 64 - 3 = 61

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