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hichkok12 [17]
3 years ago
13

I have to reflect a triangle over the line of reflection n: y = 0. What does that mean?

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
6 0

Answer:

basically just reflect it over the y-axis since it equals 0

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This is Josh’s work and solution for the equation x^2-6x-7=0:
inn [45]

Answer:

Step-by-step explanation:

the solution is right

x^2-6x-7=0:

x^2-6x-7=0 (add 7 to both sides)

x^2-6x=7

x^2-6x+9=7+9  (the coefficient of x² will be used to divide all sides)for here its 1, it will remain same ,

then we get the coefficient of x, divide it by 2 and square it and add it to both sides

which is like these

x²-6x=7

the coefficient of x is -6

-6/2 = -3, square it (-3)² = 9

then add 9 to both sides

x^2-6x+9=7+9

simplifiy the squares on the left hand side

x²+9 = (x-3)²

(x-3)^2=16

√(x-3)^2 )=±√16

x-3=± 4

x=-3±4

then simplify each sign

x=-3+4            x=-3-4

x=1                   x=-7

7 0
3 years ago
Read 2 more answers
In the figure, Δ ABC Δ XYZ. What is the perimeter of ΔABC? Show your work.
slava [35]

Answer:

49

Step-by-step explanation:

∆ABC~ ∆XYZ

AB= 11 that is half of XY = 22

therefore XZ/2 = AC (base) = 32/2 = 16

and YZ /2 = BC(hypotenuse) = 44/2 = 22

(a= 11, b= 16, c= 22)

perimeter of a triangle

P = a+b+c

P= 11+16+22= 49

8 0
3 years ago
What's the solution?
Darina [25.2K]
The answer you are looking for is c
8 0
3 years ago
Look at this expression.
Soloha48 [4]

Answer:

c 18

Step-by-step explanation:

Idid thsi in Kahn academy and got it correctly

4 0
2 years ago
Regular hexagon FGHIJK shares a common shares a common center with square ABCD on a coordinate plane .
timurjin [86]

Answer:

<h3>"The answer is the yellow lines in the attached figure. </h3>

Step-by-step explanation: As shown in the attached figure, regular hexagon FGHIJK and square ABCD shares common centre on the co-ordinate plane and AB || FG.

We are to find the line across which the combined figure will reflect onto itself.

In the attached figure, we see two lines which are yellow in colour. We can easily detect that the figure will be reflected onto itself if these two lines acts as a mirror separately.

Hence these yellow lines are the required lines."

<h3><u>Hope this helps!</u></h3>

6 0
2 years ago
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