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3 years ago

What is the factored form of the expression. 4n^2 + 24n + 36

Mathematics

1 answer:

skad [1K]3 years ago

6 0

Answer:

(2n+6)^2

Step-by-step explanation:

To solve this, you can use the given formula: x^2 + 2xy + y^2

In this case, 4n^2 is x^2, 24n is 2xy, and 36 is y^2. The next step is:

(2n)^2 + 2(2n)(6) + (6)^2

Since this equation fits into this formula ( x^2 + 2xy + y^2), we can do:

(2n+6)(2n+6) =

(2n+6)^2

Hence, the answer is (2n+6)^2

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Answer:

168.24784 m²

Step-by-step explanation:

Area of the square:

28 x 28=784 sq. m.

Area of the circle:

π(r²)=π(14)²=196π

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Area not covered by the sprinkler:

784-615.75216=168.24784 m²

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3 years ago

Read 2 more answers

Suppose you deposit $1,200 into a bank account. The account earns yearly simple interest at a rate of 1 3/4%. How many years wil

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Answer:

6 years

Step-by-step explanation:

1.75% of 1,200 = 21

so 1 year= 21

21+21= 42 2 years

42+21=63 3 years

63+21=84 4 years

81+21=105 5 years

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Or you can do

21x6=126 so it has to equal 6 years

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3 years ago

Forensic scientists use the equation h=2.6f+47.2 to estimate the height h of a woman given the length in centimeters of her femu

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Step-by-step explanation:

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4 years ago

Prove A-(BnC) = (A-B)U(A-C), explain with an example

NikAS [45]

Answer:

Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

$B = \lbrace0,\, 1 \rbrace$ .

$C = \lbrace0,\, 2 \rbrace$ .

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .

Either way, $x \in A \backslash (B \cap C)$ .

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$ , as required.

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3 years ago

If Sasha can run 720 yards in 4 minutes, how many feet can she run in 15 seconds?

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