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2 years ago

Is 0 a perfect square?

Mathematics

2 answers:

Nutka1998 [239]2 years ago

4 0

0 is a perfect square.

Because 0 * 0 = 0.

Ghella [55]2 years ago

3 0

The answer is Yes & No, depending on your definition and how you think about the question (and the answer)

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The buyer for a convenience store negotiates a single discount for a gallon of yellow shaved ice flavoring, based on the change

Komok [63]

Answer:

$8.88

Step-by-step explanation:

12.00 - 26%(3.12)= $8.88

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2 years ago

Megan has 40$ in her wallet. She wants to buy a dress. The regular price of the dress is 45$. The dress is sold at a discount of

devlian [24]

Answer:

yes

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Factor the following :

Otrada [13]

Answer:

(x - 3)(x² + 4)

Step-by-step explanation:

x³ - 3x² + 4x - 12 ( factor first/second and third/fourth terms )

= x²(x - 3) + 4(x - 3) ← factor out (x - 3 from each term

= (x - 3)(x² + 4)

8 0

2 years ago

What is the following product? <br>(4x square root 5x^2 + 2^2 square root 6)^2

tangare [24]

The product is $104 x^{4}+16 \sqrt{30} x^{4}$

Explanation:

The given expression is $\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}$

We need to determine the product of the given expression.

First, we shall simplify the given expression.

Thus, we have,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x \sqrt{5} x+2 x^{2} \sqrt{6}\right)^2$

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)^2$

Expanding the expression, we have,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)$

Now, we shall apply FOIL, we get,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}\right)^{2}+2 ( 2 x^{2} \sqrt{6})(4 x^{2} \sqrt{5})+\left(2 x^{2} \sqrt{6}\right)^{2}$

Simplifying the terms, we have,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=16 \cdot 5 x^{4}+16 \sqrt{30} x^{4}+4 \cdot 6 x^{4}$

Multiplying, we get,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=80 x^{4}+16 \sqrt{30} x^{4}+24 x^{4}$

Adding the like terms, we get,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=104 x^{4}+16 \sqrt{30} x^{4}$

Thus, the product of the given expression is $104 x^{4}+16 \sqrt{30} x^{4}$

7 0

3 years ago

the vertex of this parabola is at (5,5). when the x-value is 6, the y-value is -1. what is the coefficient of the squared in the

zlopas [31]

Answer: OPTION D.

Step-by-step explanation:

The vertex form of a quadratic function is:

$y= a(x - h)^2 + k$

Where (h, k) is the vertex of the parabola and "a" is the coefficient of the squared in the parabola's equation.

We know that the vertex of this parabola is at (5,5) and we also know that when the x-value is 6, the y-value is -1.

Then we can substitute values into $f (x) = a(x - h)^2 + k$ and solve for "a". This is:

$-1= a(6- 5)^2 + 5\\\\-1=a+5\\\\-1-5=a\\\\a=-6$

7 0

3 years ago