Ask question

Ask question

All categories

3 years ago

14

∆ABC ≅ ∆A'B'C' because you can rotate the pre-image 180° about the origin and translating it 2 units right, which is a sequence

of rigid motions. Question 1 options: True False

1 answer:

Nina [5.8K]3 years ago

6 0

Answer:

Did anyone get an answer?

Step-by-step explanation:

You might be interested in

A company's stock begins the week with a price of $43.85 per share. THe price changes by +$2.70 each day for the first two days,

mixas84 [53]

Given:

Initial price of the stock=$43.85

Change for the first two days=+$2.70

Change for next two days=-$1.10

Last day=-$4.45

The objective is to find the price at the last day.

Let's take the price at final day as <em>x</em>.

$\begin{gathered} x=43.85+2.70+2.70-1.10-1.10-4.45 \\ x=42.6 \end{gathered}$

Hence, the price of the stock at the last day is $42.6

5 0

1 year ago

Please help asap! thx

vovikov84 [41]

Answer:

y = 0

Step-by-step explanation:

The given sinusoidal equation is $y = 2 \cdot sin \left(\dfrac{\pi}{2} \cdot x + 3 \right)$

The general form of the sin function is presented as follows;

y = A·sin(B·(x - C)) + D

Where;

A = The amplitude

The period, T = 2·π/B

The frequency, f = B/2·π

C = The horizontal shift

D = The vertical shift

By comparison with the given sine function, we have;

The amplitude, A = 2

The frequency, f = B/2·π = π/2/(2·π) = 1/4

The frequency, f = 1/4 Hz

C = The horizontal shift = 3/(π/2) = 6/π

The vertical shift, D = 0

Given that the mid line of the parent function, sin(x), is the line y = 0, and that the vertical shift is 0, the midline of the function, $y = 2 \cdot sin \left(\dfrac{\pi}{2} \cdot x + 3 \right)$ , is therefore, the line;

y = 0.

3 0

3 years ago

Please help. Will mark Brainliest to whoever gets it right! :)

levacccp [35]

Answer:

A's average speed is 100. I did this same thing and passed my grade.

6 0

3 years ago

Read 2 more answers

Can someone help me pls

Simora [160]

Sorry I can not help u right now

4 0

3 years ago

Read 2 more answers

Match the expressions with their equivalent simplified expressions.

Tasya [4]

Answer:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}$

Step-by-step explanation:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}$

$\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}$

$\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}} =\sqrt[3]{\frac{(4^3)(x^{8-2})(y^{7-10})}{(5^3)}} =\sqrt[3]{\frac{4^3x^6y^{-3}}{5^3}} =\frac{4x^2y^{-1}}{5}=\frac{4x^2}{5y}$

$\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}$

$\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}$

$\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}$

Thus,

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}$

3 0

4 years ago