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alekssr [168]
4 years ago
10

Quadrilateral CDEF is reflected across the x-axis and translated 3 units left to create quadrilateral GHJK. http://prntscr.com/i

skwr8
CDEF≅GHJK

What statement about quadrilateral GHJK is true?
answers: http://prntscr.com/iskxcb

copy and paste the links if they dont work into new tabs
Mathematics
1 answer:
eimsori [14]4 years ago
7 0
To answer this question, you will reflect the original shape to make a new trapezoid at K(2, -1), G (2, -4), H(6, -4), and J (6, -2).  You will keep the y value in each ordered pair and write the opposite x value to show the reflection.

These are all then translated 3 units left to create the new ordered pairs:

K(-1, -1), G(-1, -4), H(3, -4), J(3, -2)

The two parallel sides are KG and JH.  These were originally CF and DE.
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Which of these is linear and which is non-linear?<br><br> 1. y=13x-24<br> 2. y=22x²-56
NemiM [27]

y = 13x - 24

8 0
4 years ago
Read 2 more answers
PLZ HELP ME PLZ I don't understand this
FinnZ [79.3K]

Answer:

97

Step-by-step explanation:

For this equation we have to plug in, so z= 6 and w=7. the question is 8z+7w

so plug in 6 for z and 7 for w. 8(6)+7(7) so 8*6 is 48 and 7*7= 49. 48+49= 97

hope this helps :)

5 0
3 years ago
After the booster club sold 40 hotdogs at a football game, it had $90 in profit.
podryga [215]

Step-by-step explanation:

x = goods y = $

x Sold = 40, Y = $90

x Sold = 80, Y = $210

sum of xHotdogs = 40+80 = 120 Hotdogs

Sum of Y$ = $90 + 210 = 300

so

X = 2A & Y = 3 its mean one hotdogs can sold for one each = $2.25 and we round it to $3

So = XY = 2A + 3

sorry if i wrong

4 0
3 years ago
If sin=2/3 and tan is less than 0, what is the value of cos
Sonbull [250]

tangent is less than 0 or tan(θ) < 0, is another way to say tan(θ) is negative, well, that only happens on the II Quadrant and IV Quadrant, where sine and cosine are different signs, so we know θ is on the II or IV Quadrant.

\bf sin(\theta )=\cfrac{\stackrel{opposite}{2}}{\stackrel{hypotenuse}{3}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}

\bf \pm\sqrt{3^2-2^2}=a\implies \pm\sqrt{5}=a\implies \stackrel{\textit{II Quadrant}}{-\sqrt{5}=a} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill cos(\theta )=\cfrac{\stackrel{adjacent}{-\sqrt{5}}}{\stackrel{hypotenuse}{3}}~\hfill

4 0
3 years ago
Solve the system of equations below, find the sum of x and y. 2x+4y=16 3x+y=9
STatiana [176]

y=9-3x

2x+36-12x=16

-10x=-20

x=2

y=9-3*2

y=9-6

y=3

5 0
3 years ago
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