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2 years ago

Someone help.me find angle C and Dalso this is in CIRCLE THEOREM lesson

Mathematics

2 answers:

Artyom0805 [142]2 years ago

8 0

Step-by-step explanation:

Opposite angles in a cyclic quadrilateral are supplementary.

Therefore 3c = 4d = 180°, c = 60° and d = 45°.

34kurt2 years ago

8 0

The answer is c=60 and d=45

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3 years ago

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3 years ago

Read 2 more answers

PLEASE HELP 100 POINTS!!!!!!

horrorfan [7]

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:

$g(x)=x^3-9x \quad \textsf{when x is less than 3}$

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

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.

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).

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