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Stella [2.4K]
2 years ago
7

1. Quadrilateral BGTS has vertices B (-3, 4), G (-1, 3), T (-1, 1), and S (-4, 2). Graph BGTS and its image after a translation

along (5,0) and a reflection in the x-axis.​
Mathematics
1 answer:
larisa86 [58]2 years ago
6 0

The image is attached with answer

<h3>What is a Quadrilateral ?</h3>

A quadrilateral is a polygon with four sides , four angles and four vertices.

It is given in the question that

Quadrilateral BGTS has vertices B (-3, 4), G (-1, 3), T (-1, 1), and S (-4, 2).

A graph has to be drawn of

its image after a translation along (5,0) and a reflection in the x-axis.​

The coordinates after translation and reflection is is B (8, -4), G (6, -3), T (6, -1), and S (7, -2).

The image is attached with the answer .

To know more about Quadrilateral

brainly.com/question/13805601

#SPJ1

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Hi I need help with this
ch4aika [34]

Answer:

well this is going to be long, but

1. 0.625

2. -0.136

3. 1.222

4. -5.075

5. -7.455

6. 4.067

7. -9.111

8. -7.833

9. 17/25

10. -1/100

11. -3 99/100

12. 8 149/200

13. 3 1/200

14. -13 3/250

15. -9 49/50

16. -10 113/250

17. your friends snapper is larger, 1320/156 > 1313/156

18. 0.13 > 1/8

19. -1 2/9 < 5/4

20. -5.175 < -5 1/6

21 -1 1/2 = -1.5, - 1 7/9 = -1.777777....

Step-by-step explanation:

sorry i didnt know which question you needed help on so i did it all, hope this helped you!! have a great day

4 0
3 years ago
Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
3 years ago
What integers are plotted on the number line below?
Nonamiya [84]

The correct answer would be D because if there are points at -8 and -2, those are the numbers that are plotted.

4 0
3 years ago
Read 2 more answers
What are the first three terms of a geometric sequence in which a5= 25 and the common ratio is 5?
shepuryov [24]

Answer:

1/25, 1/5, 1

Step-by-step explanation:

8 0
3 years ago
[1x-6]&gt;1 what are the solutions
a_sh-v [17]

Answer:

x > 7

(I think this is Right but not sure yet )

3 0
3 years ago
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