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bagirrra123 [75]

3 years ago

7

Vertex A in quadrilateral ABCD lies at (-3, 2). If you rotate ABCD 180° clockwise about the origin, what will be the coordinates

of A′ of the rotated quadrilateral A′B′C′D′?

1 answer:

natali 33 [55]3 years ago

4 0

The rule for a rotation by 180° about the origin is (x,y) -->(−x,−y)

So A(-3, 2) to A'(3, -2)

Answer:

A'(3, -2)

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The slope is -1/3

Step-by-step explanation:

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-1/3

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3 years ago

IM 20 MIN AWAY FROM FAILING PLZZZ HELPPPP

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Answer:

13 and 32

Step-by-step explanation:

let one number be x

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3 years ago

A post is supported by two wires (one on each side going in oppositedirections) creating an angle of 80° between the wires. The

Vladimir [108]

Using the Sine rule,

$\frac{\sin A}{A}=\frac{\sin B}{B}=\frac{\sin C}{C}$ $\begin{gathered} \text{Let A = 14m,} \\ Substituting the variables into the formula,Where the length of the wires are, AP = xm and BP = ym[tex]\begin{gathered} \frac{\sin80^0}{14}=\frac{\sin40^0}{x} \\ \text{Crossmultiply,} \\ x\times\sin 80^0=14\times\sin 40^0 \\ Divide\text{ both sides by }\sin 80^0 \\ x=\frac{14\sin40^0}{\sin80^0} \\ x=9.14m \end{gathered}$

Hence, the length of wire AP (x) is 9.14m.

For wire BP (y)m,

Sum of angles in a triangle is 180 degrees,

$A^0+P^0+B^0=180^0$ $\begin{gathered} \text{Where A}^0=\text{ unknown,} \\ P^0=80^0\text{ and,} \\ B^0=40^0 \\ A^0+80^0+40^0=180^0 \\ A^0+120^0=180^0 \\ A^0=180^0-120^0 \\ A^0=60^0 \end{gathered}$

Using the side rule to find the length of wire BP,

$\begin{gathered} \frac{\sin 60^0}{y}=\frac{\sin 80^0}{14} \\ \text{Crossmultiply,} \\ y\times\sin 80^0=14\times\sin 60^0 \\ \text{Didive both sides by }\sin 80^0 \\ y=\frac{14\times\sin 60^0}{\sin 80^0} \\ y=12.31m \end{gathered}$

Hence, the length of wire BP (y) is 12.31m

Therefore, the length of the wires are (9.14m and 12.31m).

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