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beks73 [17]
2 years ago
12

PLEASE HELP 100 POINTS!!!!!!

Mathematics
1 answer:
horrorfan [7]2 years ago
3 0

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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Step-by-step explanation:

The original ratio was 7:1. So, that means if you added another part to the salt it would make 2. That means that the 7 parts of water would become 14 parts. Since the ratio of the salt part was doubled from 1 to 2, you would have to do the same with the water ratio part. Hoped this helped!

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Help me plz I’m struggling with this
elena-s [515]

Answer:

AC = 8√3       AC = 7.65          

A = 43.17°       AB = 16.8

B = 46.83°      B = 27°

Step-by-step explanation:

FIRST TRIANGLE

by using pythagorus theorem:

Hypo² = Base² + height²

19² = 13² + AC²

AC² = 19² - 13²

AC² = 192

AC = √192

AC = 8√3

sinФ =base/hypo

sin A = 13/19

A = sin^-1 (13/19)

A = 43.17°

43.17°+ B + 90° =180 (sum of angles of triangle)

B = 180° - 133.17°

  = 46.83°

<h3>SECOND TRIANGLE</h3>

TanФ = base/height

Tan 63° = 15 / AC

1.96 = 15/AC

AC = 15/1.96

AC = 7.65

AB² = AC² + BC²

AB² = 7.65² + 15²

AB² = 283.5

AB = √283.5

AB = 16.8

tan B = AC /BC

tanФ = 7.65/15

tanФ = 0.51

Ф = tan^-1(0.51)

B = 27°

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3 years ago
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