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Elena L [17]
3 years ago
13

The rule is applied to ΔABC. On a coordinate plane, 5 triangles are shown. Triangle A B C has points (2, negative 4), (4, negati

ve 4), (4, negative 2). Triangle 1 has points (negative 4, 2), (negative 4, 4), (negative 2, 4). Triangle 2 has points (negative 3, 3), (negative 1, 3), (negative 1, 1). Triangle 3 has points (1, 1), (1, 3), (3, 3). Triangle 4 has points (2, 4), (4, 2), (4, 4). Which triangle shows the final image? 1 2 3 4
Mathematics
2 answers:
vfiekz [6]3 years ago
8 0

Answer:

Consider triangle ABC with vertices at points A(2,-4), B(4,-4) and C(4,-2).

1. The rotation  acts with the rule:

Then:

2. The reflection across the y-axis has a rule:

So,

Triangle A''B''C'' is exactly the same as tiangle from figure 1.

Answer: correct choice is 1.

PLZ brainliest answer

Step-by-step explanation:

Luden [163]3 years ago
5 0

Answer:

A(2,-4), B(4,-4) and C(4,-2).

Step-by-step explanation:

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Question: Find The Sum Of The Integers From -6 To 58 Of Ss. O 1,500 O 1,734 O 1,690 O 1,621​
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Answer:

The Sum Of The Integers From -6 To 58 is <u>1690.</u>

Step-by-step explanation:

Given,

a=-6

T_n=58

We have to find out the sum of integers from -6 To 58.

Firstly we will find out the total number of terms that is 'n'.

Here a_1=-6\ and\ a_2=-5

\therefore d=a_2-a_1=-5-(-6)=-5+6=1

Now we use the formula of A.P.

T_n=a+(n-1)d

On substituting the values, we get;

58=-6+(n-1)1\\\\n-1=58+6\\\\n-1=64\\\\n=64+1=65

So there are 65 terms in between  -6 To 58.

That means we have to find the sum of 65 terms in between  -6 To 58.

Now we use the formula of Sum of n_terms.

S_n=\frac{n}{2}(2a-(n-1)d)

On substituting the values, we get;

S_{65}=\frac{65}{2}(2\times-6+(65-1)1)\\\\S_{65}=\frac{65}{2}(-12+64)\\\\S_{65}=\frac{65}{2}\times52\\\\S_{65}=65\times26=1690

Hence The Sum Of The Integers From -6 To 58 is <u>1690.</u>

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