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slega [8]
2 years ago
9

Point HH is located at (4,-2)(4,−2) on the coordinate plane. Point HH is reflected over the yy-axis to create point H'H ′ . Poin

t H'H ′ is then reflected over the xx-axis to create point H''H ′′ . What ordered pair describes the location of H''?H ′′ ?
Mathematics
1 answer:
marin [14]2 years ago
4 0

Answer: Coordinates of H  : (−4,2)

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Solve.<br> 4/5x-33/4x=21/2<br> A. -1 9/50<br> B. -50/59<br> C.34/75<br> D.19/50
vaieri [72.5K]
I think the answer is choice D, but don't trust me I didn't do the math.
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Y varies directly with x. When x is 4 y is 8. Find y when x = 10.
grandymaker [24]

Step-by-step explanation:

y = 4 when x = 8 So, 4 = 8k

k = 1/2

y = (1/2)x

Therefore, when x = 10, y = (1/2)(10) = 5

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Q7Which statement is incorrect for the graph of thefunction y = -5 cosGcos( 3 (x – 3)) +62The period is 4.The amplitude is 3.The
guajiro [1.7K]

The given function is,

y=-5\cos (\frac{\pi}{2}(x-3))+6

The graph can be drawn as,

The period will be distance between two consequetive maximas,

\text{Period}=1-(-3)=4

Thus, period is correct.

The amplitude can be determined as,

\begin{gathered} Amplitude=\frac{\max -\min }{2} \\ =\frac{11-1}{2} \\ =5 \end{gathered}

Thus, amplitue is incorrect.

The range is,

y\in\lbrack1,11\rbrack

Thus, the range is correct.

The midline is,

\begin{gathered} \text{midline}=\frac{\max +\min }{2} \\ =\frac{1+11}{2} \\ =6 \end{gathered}

Thus, the midline is correct.

Thus, option (b) is the solution.

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1 year ago
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The answer of above equation is f
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3 years ago
Triangle A″B″C″ is formed by a reflection over x = 1 and dilation by a scale factor of 2 from the origin. Which equation shows t
andrew-mc [135]

Answer:

segment AB over segment A double prime B double prime = the square root of 13 over 2 times the square root of 13

Step-by-step explanation:

Triangle ABC has vertices at points A(-3,3), B(1,-3) and C(-3,-3).

1. Reflection over x = 1 maps vertices A, B and C as follows

  • A(-3,3)→A'(5,3);
  • B(1,-3)→B'(1,3);
  • C(-3,-3)→C'(5,-3).

2. Dilation by a scale factor of 2 from the origin has the rule

(x,y)→(2x,2y)

So,

  • A'(5,3)→A''(10,6);
  • B'(1,3)→B''(2,6);
  • C'(5,-3)→C''(10,-6)

See attached diagram for details

Note that

A''B''=2AB\\ \\A''C''=2AC\\ \\B''C''=2BC,\\ \\AB=\sqrt{(-3-1)^2+(3-(-3))^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}

so

\dfrac{AB}{A''B''}=\dfrac{2\sqrt{13}}{2\cdot 2\sqrt{13}}=\dfrac{\sqrt{13}}{2\sqrt{13}}

7 0
3 years ago
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