“huh” is the interjection!
<span> The product of two perfect squares is a perfect square.
Proof of Existence:
Suppose a = 2^2 , b = 3^2 [ We have to show that the product of a and b is a perfect square.] then
c^2 = (a^2) (b^2)
= (2^2) (3^2)
= (4)9
= 36
and 36 is a perfect square of 6. This is to be shown and this completes the proof</span>
Answer:
x = 8
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality
<u>Algebra I</u>
Step-by-step explanation:
<u>Step 1: Define</u>
3(x - 2) + 8 = 2(x + 5)
<u>Step 2: Solve for </u><em><u>x</u></em>
- (Parenthesis) Distribute: 3x - 6 + 8 = 2x + 10
- Combine like terms: 3x + 2 = 2x + 10
- {Subtraction Property of Equality] Subtract 2x on both sides: x + 2 = 10
- [Subtraction Property of Equality] Subtract 2 on both sides: x = 8
<u>Step 3: Check</u>
<em>Plug in x into the original equation to verify it's a solution.</em>
- Substitute in <em>x</em>: 3(8 - 2) + 8 = 2(8 + 5)
- (Parenthesis) Subtract/Add: 3(6) + 8 = 2(13)
- Multiply: 18 + 8 = 26
- Add: 26 = 26
Here we see that 26 does indeed equal 26.
∴ x = 8 is the solution to the equation.
Yes 0.634 is a rational
number.
Rational numbers are those numbers that can be still expressed in
standard form or in fraction form and vice-versa. Unlike irrational numbers
that are opposed to the definition of rational numbers. These values include
pi, square root of two and etc. These values are impossible to fractionize.
To better illustrate this
circumstance.
We can have calculate a number that will have a quotient of 0.634
or a fraction that is equal to the given value.
<span><span>
1. </span><span> 634/1000 =
0.634</span></span>
<span><span>2. </span><span> 317/500 = 0.634
</span></span>
3/4 ÷ 1/5
= 3/4 × 5/1
= 15/4
i am a mathematics teacher. if anything to ask please pm me
