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Korvikt [17]
2 years ago
8

Trapezoid EFGH has coordinates E(−3, 4) , F(1, 4), G(3, 1) , and H(−5, 1) . Trapezoid E'F'G'H' has coordinates E′(−12, 16) , F′(

4, 16), G'(12, 4) , and H′(−20, 4) . Trapezoid E'"F"G"H" has coordinates E′′(12, −16), F′′(−4, −16), G′′(−12, −4) , and H′′(20, −4) .
Which transformations describe why trapezoids EFGH and E'"F"G"H" are similar?

A.​ Trapezoid EFGH ​ was translated 4 units right and 4 units up and then rotated 180° clockwise.

​ B. Trapezoid EFGH ​ was rotated 90° clockwise and then dilated by a scale factor of 4.

C.​ Trapezoid EFGH ​ was dilated by a scale factor of 4 and then rotated 180° counterclockwise.

​D. Trapezoid EFGH ​ was dilated by a scale factor of 14 and then reflected across the x-axis.
Mathematics
2 answers:
julia-pushkina [17]2 years ago
8 0
I believe that the answer would be C
Elan Coil [88]2 years ago
5 0

Answer Very much Correct i took the test

Step-by-step explanation:

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Find the mean, median and mode of the weights of the people shown. 105kg 53kg 76kg 91kg 120kg 61kg 55kg 98kg 61kg
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First, you need to put them in order
53kg, 55kg, 61kg, 61kg, 76kg, 91kg, 98kg, 105kg, 120kg

For mean, you add them all up and divide by the amount of numbers (9)
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For median, you find the middle number (76kg)

For mode, you find the number that appears the most (61kg)

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2 years ago
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R+5/mn=p solve for m
AnnZ [28]

Answer:

               \bold{m\ =\ \dfrac5{(p-r)n}}

Step-by-step explanation:

                                             \bold{r+\dfrac5{mn}\ =\ p}\\\\ {}\quad-r\qquad-r\\\\{}\ \ \bold{\dfrac5{mn}\ =\ p-r}\\\\{}\ ^{_\times}(mn)\quad ^{_\times}(mn)\\\\{}\quad\bold{5\ =\ (p-r)^{_\times}(mn)}\\\\\div(p-r)\quad\div(p-r)\\\\{}\ \ \bold{\dfrac5{p-r}\ =\ mn}\\\\{}\quad \ \div n\quad\ \ \div n\\\\\bold{\dfrac5{(p-r)n}\ =\ m}

If you mean (r+5)/mn then:

\bold{\dfrac{r+5}{mn}\ =\ p}\\\\{}\ ^{_\times}(mn)\quad ^{_\times}(mn)\\\\{}\ \bold{r+5\ =\ pmn}\\\\\div(pn)\quad\div(pn)\\\\{}\ \ \bold{\dfrac{r+5}{pn}\ =\ m}

4 0
2 years ago
Emery looks out their apartment window to the building across the way. The building isknown to be 42 feet tall. The angle of dep
Amanda [17]

Answer:

x=44.5ft

Step-by-step explanation:

From the question we are told that:

Height h=42ft

Angle of depression \theta=27\textdegree

Angle of Elevation \alpha=23 \textdegree

Generally the equation for the vertical distance between Emery's distance x and the bottom of the building  is mathematically given by

Since the angle of depression and elevation are given as

27 and 23 respectively

Therefore

Emery's view of the 42 ft building is

\gamma=23+27

\gamma=50  \textdegree

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h'=\frac{27}{50}*42

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x=\frac{h'}{tan\theta}

x=\frac{22.68}{tan 27}

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