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Serga [27]
3 years ago
15

HELP!! Identify the transformation from ABC to A'B'C'.

Mathematics
1 answer:
elena55 [62]3 years ago
7 0

Answer:

<em>Thus, the transformation from ABC to A'B'C' is a reflection over the x-axis.</em>

<em>Choice 1.</em>

Step-by-step explanation:

<u>Reflection over the x-axis</u>

Given a point A(x,y), a reflection over the x-axis maps A to the point A' with coordinates A'(x,-y).

The figure shows triangles ABC and A'B'C'. It can be clearly seen the x-coordinates for each vertex of both triangles is the same and the y-coordinate is the inverse of it counterpart. For example A=(5,3) and A'=(5,-3)

Thus, the transformation from ABC to A'B'C' is a reflection over the x-axis.

Choice 1.

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Last month, Grayson and Justine sold candy to raise money for their debate team. Justine sold 1 1/5 times as much candy as Grays
ale4655 [162]

Answer: 4/5 of a box of candy

Step-by-step explanation:

Justine sold 1¹/₅ times the ²/₃ of a box of candy that Grayson sold.

Justin sold;

= 1¹/₅ * ²/₃

= 6/5 * 2/3

= 12/15

= 4/5 of a box of candy

6 0
3 years ago
Really need help , it will help my grade!!!
klio [65]

Answer:

ax + b= ax + b

Step-by-step explanation:

Infinite many solutions when both sides are equal

so

ax + b= ax + b

5 0
3 years ago
WORTH 20 POINTS PLEASE HELP
kap26 [50]
I think its D but again I ight be wrong.
4 0
3 years ago
Please help I really need it​
yawa3891 [41]

Answer: m< MIH is 34°

M< AVM is 70°

And the angle of the obtuse angle formed at the intersection of AV and HI is 104°

Step-by-step explanation: starting with m<MIH, AH is parallel to MI, so that would make the same angle H has the same for I. (34°)

Next is m<AVM. The angle of m<LAH is 110°. So the angle of m<HAV (because it's supplementary of it) is 70 (110+70=180). Which makes m<AVM 70° since it's vertical to each other.

Since we got those answers, the next one you just plug it in, and the answer would be 104°

5 0
3 years ago
{20,1,6,10,11}<br> Find the subset and the proper subset.
liubo4ka [24]
With 5 elements in A={20,1,6,10,11}, there are 2^5=32 possible subsets, including
the null set, and A itself.
Any subset that is identical to A is NOT a proper subset.
Therefore there are 31 proper subsets, plus the subset {20,1,6,10,11}.

The subsets are:
null set  {} (has no elements) ........total 1
{20},{1},{6},{10},{11}.......................total 5
{20,1},{20,6}...{10,11}.....................total 10
{20,1,6},{20,1,10},...{6,10,11}.........total 10
{20,1,6,10}...{1,6,10,11}.................total 5
{20,1,6,10,11}.................................total 1
Altogether 32 subsets.


4 0
3 years ago
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