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3 years ago

What is a square root of a number

1 answer:

3 0

Finding the square root of a number is the inverse operation of squaring that number. Remember, the square of a number is thatnumber times itself. The perfect squares are the squares of the wholenumbers. The square root of a number, n, written below is the number that gives n when multiplied by itself.

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Need to finish this today ....

lesya692 [45]

The correct answer is A

3 0

3 years ago

Read 2 more answers

At Prairieview Middle School, 100 students participate in at least one of three after-school clubs: soccer club, art club, and g

Tresset [83]

Answer:

Step-by-step explanation:

Number of students in soccer club, n(S) = 50

Number of students in Art club, n(A) = 53

Number of students in Gaming club, n(G)

n( $S \cup A\cup G$ ) = 100

n( $S \cap A\cap G$ ) = 9

n( $S \cap A$ ) = 20

n( $A \cap G$ ) = 35

n( $G \cap S$ ) = 29

Formula:

n ( A ∪ B ∪ C ) = n(A) + n(B) + n(C) – n ( A ∩ B ) – n(B ∩ C) – n (A ∩ C) + n( A ∩ B ∩ C )

Putting the values:

100 = 50 + 53 + n(G) - 20 - 35 - 29 + 9

100 = 112 + n(G) - 84

n(G) = 72

Number of students in gaming club only = n(G) - n( $G \cap S$ ) - n( $A \cap G$ ) + n( $S \cap A\cap G$ )

= 72 - 35 - 29 + 9

= <em>17</em>

6 0

3 years ago

Factor the polynomial, x2 + 5x + 6

patriot [66]

Answer:

Choice b.

$x^{2} + 5\, x + 6 = (x + 3)\, (x + 2)$ .

Step-by-step explanation:

The highest power of the variable $x$ in this polynomial is $2$ . In other words, this polynomial is quadratic.

It is thus possible to apply the quadratic formula to find the "roots" of this polynomial. (A root of a polynomial is a value of the variable that would set the polynomial to $0$ .)

After finding these roots, it would be possible to factorize this polynomial using the Factor Theorem.

Apply the quadratic formula to find the two roots that would set this quadratic polynomial to $0$ . The discriminant of this polynomial is $(5^{2} - 4 \times 1 \times 6) = 1$ .

$\begin{aligned}x_{1} &= \frac{-5 + \sqrt{1}}{2\times 1} \\ &= \frac{-5 + 1}{2} \\ &= -2\end{aligned}$ .

Similarly:

$\begin{aligned}x_{2} &= \frac{-5 - \sqrt{1}}{2\times 1} \\ &= \frac{-5 - 1}{2} \\ &= -3\end{aligned}$ .

By the Factor Theorem, if $x = x_{0}$ is a root of a polynomial, then $(x - x_0)$ would be a factor of that polynomial. Note the minus sign between $x$ and $x_{0}$ .

The root $x = -2$ corresponds to the factor $(x - (-2))$ , which simplifies to $(x + 2)$ .
The root $x = -3$ corresponds to the factor $(x - (-3))$ , which simplifies to $(x + 3)$ .