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jenyasd209 [6]
3 years ago
6

Figure ABCDEF was reflected across the line y = –x to create figure A'B'C'D'E'F'.

Mathematics
2 answers:
Molodets [167]3 years ago
3 0

Answer:

(2, –4)

Step-by-step explanation:

just took it on e2020

VladimirAG [237]3 years ago
3 0

Answer:

Its (2,-4)

Step-by-step explanation:

Trust you took a test with it

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I need the work shown not just the answers please
Yuliya22 [10]
1.) To find the surface area, we want to find the area of each side, then add them all up together:
Area of base=lwA=(5)(5)A=25 cm²
Area of triangle=1/2bhA=(1/2)(5)(12.5)A=31.25 cm²
Since the 4 triangles have the same area, we can multiply the area we found by 4 and then add the area of the square:
SA=(4)(31.25) + 25SA=125+25SA=150 cm²

2.) <span>To find the volume use the formula: 

V=(1/3)lwh
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3 years ago
What is the missing reason in the proof shown ?
vodomira [7]

Answer:

A symmetric property

Step-by-step explanation:

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Nathaniel bought headphones online for $22. He used a coupon code to get a 40% discount. The website also applied a 10% processi
Tatiana [17]

Answer:

8.80 if it's the 40°/.

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8 0
2 years ago
Factor 9x6 – 16 y completely.
sveticcg [70]

Answer:

(3x^3 - 4y^3) ( 3x^3 + 4y^3)

Step-by-step explanation:

9x^6 – 16 y^6

Rewriting as

(3x^3) ^2 - ( 4y^3) ^2

This is the difference of squares a^2 - b^2 = (a-b)(a+b)

(3x^3 - 4y^3) ( 3x^3 + 4y^3)

6 0
3 years ago
Find the parametric equations for the line that is tangent to the given curve at the given parameter (3cost)i (t^4 -3sint)j
ale4655 [162]

Answer:

Following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

In the given equation, when the point t=0

So,

\to r(0) = (3 \cos 0)i + (0^4 - 6 \sin 0)j + (2e^{3\times 0})k)

           = (3 \times 1)i + (0 - 0)j + (2e^{0})k)\\\\ = 3i +  0j + (2 \times 1)k)\\\\ = 3i +  0j + 2k \\

The value of the coordinates are 3, 0, 2 . so, the equation of the line is:

\to \frac{(x-3)}{3 \cos \ t} = \frac{(y-0)}{(t^{4}-6 \sin \ t)} = \frac{(z-2)}{2e^{3t}}=k

5 0
3 years ago
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