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Marat540 [252]
3 years ago
13

HELP 5 STAR + THANKS What series of transformations from △ABC to ​ △DEF ​ shows that △ABC≅△DEF ? a reflection across the y-axis

followed by a translation of 1 unit right and 2 units up a clockwise rotation of 90° about the origin followed by a translation of 4 units right and 4 units up a reflection across the x-axis followed by a translation of 1 unit right and 1 unit down a reflection across the line y = x followed by a positive rotation of 270° about the center
Mathematics
1 answer:
nirvana33 [79]3 years ago
7 0
In the given triangle, the verteces are A(-4, 1), B(-6, 5), C(-1, 2).
A refrection across the x-axis will result in A'(-4, -1), B'(-6, -5), C'(-1, -2)
A translation of 1 unit to the right will result in A"(-3, -1), B"(-5, -5), C"(0, -2)
A translation of 1 unit down will result in A"'(-3, -2), B"'(-5, -6), C"'(0, -3) which corresponds to points DEF.
Therefore, the series of transformation required to transform ABC to DEF are <span>a reflection across the x-axis followed by a translation of 1 unit right and 1 unit down.</span>
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What is the true solution to the logarithmic equation?<br>log2 [log2(sqrt4x)]=1​
erastova [34]

Answer:4

Step-by-step explanation:

log₂[log₂(√4x)] = 1

log₂2 =1

So we replace our 1 with log₂2

log₂[log₂(√4x)] = log₂2

log₂ on bothside will cancel each other.

We will be left with;

[log₂(√4x)] = 2

log = power of exponential

√4x = 2²

√4x = 4

Square bothside

(√4x)² = 4²

4X = 16

Divide bothside by 4

4x/4 = 16/4

x = 4

8 0
3 years ago
Read 2 more answers
To ___ is to use place value to exchange equal amounts when renaming a number.
Vanyuwa [196]
To regroup is to use place value to exchange equal amount when renaming a number.
6 0
3 years ago
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Simplify u^2+3u/u^2-9<br> A.u/u-3, =/ -3, and u=/3<br> B. u/u-3, u=/-3
VashaNatasha [74]
  The correct answer is:  Answer choice:  [A]:
__________________________________________________________
→  "\frac{u}{u-3} " ;  " { u \neq ± 3 } " ; 

          →  or, write as:  " u / (u − 3) " ;  {" u ≠ 3 "}  AND:  {" u ≠ -3 "} ; 
__________________________________________________________
Explanation:
__________________________________________________________
 We are asked to simplify:
  
  \frac{(u^2+3u)}{(u^2-9)} ;  


Note that the "numerator" —which is:  "(u² + 3u)" — can be factored into:
                                                      →  " u(u + 3) " ;

And that the "denominator" —which is:  "(u² − 9)" — can be factored into:
                                                      →   "(u − 3) (u + 3)" ;
___________________________________________________________
Let us rewrite as:
___________________________________________________________

→    \frac{u(u+3)}{(u-3)(u+3)}  ;

___________________________________________________________

→  We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ;  since:

" \frac{(u+3)}{(u+3)} = 1 "  ;

→  And we have:
_________________________________________________________

→  " \frac{u}{u-3} " ;   that is:  " u / (u − 3) " ;  { u\neq 3 } .
                                                                                and:  { u\neq-3 } .

→ which is:  "Answer choice:  [A] " .
_________________________________________________________

NOTE:  The "denominator" cannot equal "0" ; since one cannot "divide by "0" ; 

and if the denominator is "(u − 3)" ;  the denominator equals "0" when "u = -3" ;  as such:

"u\neq3" ; 

→ Note:  To solve:  "u + 3 = 0" ; 

 Subtract "3" from each side of the equation; 

                       →  " u + 3 − 3 = 0 − 3 " ; 

                       → u =  -3 (when the "denominator" equals "0") ; 
 
                       → As such:  " u \neq -3 " ; 

Furthermore, consider the initial (unsimplified) given expression:

→  \frac{(u^2+3u)}{(u^2-9)} ;  

Note:  The denominator is:  "(u²  − 9)" . 

The "denominator" cannot be "0" ; because one cannot "divide" by "0" ; 

As such, solve for values of "u" when the "denominator" equals "0" ; that is:
_______________________________________________________ 

→  " u² − 9 = 0 " ; 

 →  Add "9" to each side of the equation ; 

 →  u² − 9 + 9 = 0 + 9 ; 

 →  u² = 9 ; 

Take the square root of each side of the equation; 
 to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ; 

→ √(u²) = √9 ; 

→ | u | = 3 ; 

→  " u = 3" ; AND;  "u = -3 " ; 

We already have:  "u = -3" (a value at which the "denominator equals "0") ; 

We now have "u = 3" ; as a value at which the "denominator equals "0"); 

→ As such: " u\neq 3" ; "u \neq -3 " ;  

or, write as:  " { u \neq ± 3 } " .

_________________________________________________________
6 0
3 years ago
Name two pairs of adjacent angles and two pairs of vertical angles in the figure
insens350 [35]
Adjacent angles :

IGJ and IGH

KGL and LGM

vertical angles :

JGK and HGM

IGJ and MGL


hope this helps

7 0
3 years ago
Please solve with explanation
Juliette [100K]

Step-by-step explanation:

4\frac{9}{20} - 1\frac{2}{6} =\\\frac{89}{20}  - \frac{8}{6} = \frac{89 - 26.7}{20} = \frac{62.3}{20} \\

4\frac{6}{9}  - 3\frac{3}{6} =\\\frac{42}{9} - \frac{21}{6} = \frac{252 - 189}{54} = \frac{63}{54} = 1\frac{9}{54}

2\frac{2}{4} - 1\frac{1}{3} =\\\frac{10}{4} - \frac{4}{3} = \frac{30 - 16}{12} = \frac{14}{12} = 1\frac{2}{12}

8 0
4 years ago
Read 2 more answers
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