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aliya0001 [1]
3 years ago
9

A woman is looking for a bigger square office. She finds an office twice the area of her current office. If the perimeter of her

current office is 88 feet, how many square feet is the new office?
Mathematics
2 answers:
katovenus [111]3 years ago
5 0

Answer:

The area of the new office = 968 square feet.

Step-by-step explanation:

Given: A woman is looking for a bigger square office. She finds an office twice the area of her current office.

Let "x" be the side of length of the current office. Which is in shape of square.

A square has four sides and all the sides are in equal measure.

Perimeter of the current office = 4x

Given: Perimeter = 88 feet

Now let's find the side of the current office.

88  = 4x

Divided both sides, we get

4x/4 = 88/4

x = 22 feet.

So the side of the current office is 22 feet.

Now let's find the area of the current office = side *side

= 22*22

The area of the current office = 484 square feet.

We need to find the area of the new office, which is twice the area of the current office."

The area (square feet) of the new office = 2*484 = 968 square feet.

Sauron [17]3 years ago
4 0
88×2=176this is the answer
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The domain the given graph is :

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Answer:

A)  See attached for graph.

B)  (-3, 0)  (0, 0)  (18, 0)

C)   (-3, 0) ∪ [3, 18)

Step-by-step explanation:

Piecewise functions have <u>multiple pieces</u> of curves/lines where each piece corresponds to its definition over an <u>interval</u>.

Given piecewise function:

g(x)=\begin{cases}x^3-9x \quad \quad \quad \quad \quad \textsf{if }x < 3\\-\log_4(x-2)+2 \quad  \textsf{if }x\geq 3\end{cases}

Therefore, the function has two definitions:

  • g(x)=x^3-9x \quad \textsf{when x is less than 3}
  • g(x)=-\log_4(x-2)+2 \quad \textsf{when x is more than or equal to 3}

<h3><u>Part A</u></h3>

When <u>graphing</u> piecewise functions:

  • Use an open circle where the value of x is <u>not included</u> in the interval.
  • Use a closed circle where the value of x is <u>included</u> in the interval.
  • Use an arrow to show that the function <u>continues indefinitely</u>.

<u>First piece of function</u>

Substitute the endpoint of the interval into the corresponding function:

\implies g(3)=(3)^3-9(3)=0 \implies (3,0)

Place an open circle at point (3, 0).

Graph the cubic curve, adding an arrow at the other endpoint to show it continues indefinitely as x → -∞.

<u>Second piece of function</u>

Substitute the endpoint of the interval into the corresponding function:

\implies g(3)=-\log_4(3-2)+2=2 \implies (3,2)

Place an closed circle at point (3, 2).

Graph the curve, adding an arrow at the other endpoint to show it continues indefinitely as x → ∞.

See attached for graph.

<h3><u>Part B</u></h3>

The x-intercepts are where the curve crosses the x-axis, so when y = 0.

Set the <u>first piece</u> of the function to zero and solve for x:

\begin{aligned}g(x) & = 0\\\implies x^3-9x & = 0\\x(x^2-9) & = 0\\\\\implies x^2-9 & = 0 \quad \quad \quad \implies x=0\\x^2 & = 9\\\ x & = \pm 3\end{aligned}

Therefore, as x < 3, the x-intercepts are (-3, 0) and (0, 0) for the first piece.

Set the <u>second piece</u> to zero and solve for x:

\begin{aligned}\implies g(x) & =0\\-\log_4(x-2)+2 & =0\\\log_4(x-2) & =2\end{aligned}

\textsf{Apply log law}: \quad \log_ab=c \iff a^c=b

\begin{aligned}\implies 4^2&=x-2\\x & = 16+2\\x & = 18 \end{aligned}

Therefore, the x-intercept for the second piece is (18, 0).

So the x-intercepts for the piecewise function are (-3, 0), (0, 0) and (18, 0).

<h3><u>Part C</u></h3>

From the graph from part A, and the calculated x-intercepts from part B, the function g(x) is positive between the intervals -3 < x < 0 and 3 ≤ x < 18.

Interval notation:  (-3, 0) ∪ [3, 18)

Learn more about piecewise functions here:

brainly.com/question/11562909

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