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4 years ago

HELP WITH GEOMETRY!!

2 answers:

6 0

Similarity transformation is when you do a transformation but keep PROPOTIOALITY

The sides have to have the same scale factor with the other shapes corrisponding sides and the angles have to be ridged(have same angle measure).

otez555 [7]4 years ago

6 0

Answer:

Similarity transformations include dilation and any number of rigid transformations. In a similarity transformation, the ratio of the lengths of corresponding sides in the image and the preimage is equal to the scale factor of dilation. Two or more similar figures will have corresponding sides that are proportional in length.

Step-by-step explanation:

plato

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A model length of 12cm. Represents an actual length of 102 ft. What is the scale for the model

stealth61 [152]

102 ft = 12 cm1224 in = 12 cm1 in = 0.00980 cm

8 0

3 years ago

The domain of f(x) is the set of all real values except 7 , and the domain of g(x) is the set of all real values except -3. Witc

Anna007 [38]

Answer: <span><span>the domain of g [f(x) ] is the set of all real values except 7 and the x for which f(x) = - 3.</span>

Explanation:

Taking (g•f)(x) as (g o f) (x), this is g (x) composed with f(x) you have this analysis.

(g o f) (x) is g [ f(x) ], which means that you first apply the function f and then apply the function g to the output of f(x).

The domain of g [ f(x) ] has to exclude 7, because it is not included in the domain of f(x).

Also the domain thas to exclude those values of x for which f(x) is - 3, because the domain of g(x) is the set of all real values except - 3.

So, the domain of g [f(x) ] is the set of all real values except 7 and the x for which f(x) = - 3.

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6 0

3 years ago

Read 2 more answers

Pls help and find the area

Zepler [3.9K]

Answer:

I don’t know the area but here’s a hint to help you (hint: length x with x height)

Step-by-step explanation:

5 0

3 years ago

Can someone help me with this? PLs i'm so confused!

barxatty [35]

1. E. $sine\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{5}{13}$

2. L. $cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{12}{13}$

3. $tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{5}{12}$

4. Y. $sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{5}{13}$

5. W. $cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{12}{13}$

6. $tan\ B = \frac{b}{a} = \frac{adjacent}{opposite} = \frac{AC}{BC} = \frac{12}{5} = 2\frac{2}{5}$

7. $sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{2}$

8. W. $cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{\sqrt{3}}{2}$

9. I. $tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

10. $sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{2}$

11. E. $cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{\sqrt{3}}{1} = \sqrt{3}$

12. I. $tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

13. U. $sin\ A = \frac{a}{c} = \frac{hypotenuse}{opposite} = \frac{BC}{AB} = \frac{12}{15} = \frac{4}{5}$

14. I. $cos\A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{9}{15} = \frac{3}{5}$

15. $tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{12}{9} = \frac{4}{3} = 1\frac{1}{3}$

16. R. $sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{4}{\sqrt{65}} = \frac{4}{\sqrt{65}} * \frac{\sqrt{65}}{\sqrt{65}} = \frac{4\sqrt{65}}{65}$

17. M. $cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{7}{4} = 1\frac{3}{4}$

18. N. $tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{4}{7}$

19. L. $sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{16}{34} = \frac{8}{17}$

20. H. $cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \fac{AC}{AB} = \frac{30}{34} = \frac{15}{17}$

21. $tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{16}{30} = \frac{8}{15}$

22. O. $sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

23. O. $cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

24. N. $tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{1} = 1$

7 0

3 years ago

Help please will give brainliest

dimulka [17.4K]

Answer:

the black points on the top right marked with the letters are the answers

Hope it Helps

4 0

3 years ago