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3 years ago

Please help ASAP

1 answer:

4 0

It should be multiplied by the opposite of the coefficient of y in the first equation, hence
B)) -2

_____
Actually, it depends on your choice of next step. Multiplying by -2 lets you add the two equations. Multiplying by 2 means you need to subtract one equation from the other. We usually find addition easier, so we would usually choose -2, but either can work. It depends on whether you like to add negative numbers or subtract positive ones.

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What is the product of 4xy and y2 + 2x?

juin [17]

The product of 4xy and y² + 2x can also be written as
4xy(y² + 2x)

When we have brackets like this, we multiply whatever is left of the brackets by everything inside.

1) 4xy multiplied by y²
    4xy x y² = 4xy³

2) 4xy multiplied by 2x
    4xy x 2x = 8x²y

3) Add together 4xy³ and 8x²y
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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.

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frosja888 [35]

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