Answer: 9(3 + 2r)
<u>Step-by-step explanation:</u>
27 18r
∧ ∧
<u>3</u> 9 <u>2</u> 9r
∧ ∧
<u>3</u> <u>3</u> <u>3</u> <u>3r</u>
27: 3 * 3 * 3
18r: 2 * 3 * 3 * r
Common Factors are 3 * 3 = 9
27: 9 * <u>3</u>
18: 9 * <u>2r</u>
27 + 18r = 9(3 + 2r)
Answer:
No, to be a function a relation must fulfill two requirements: existence and unicity.
Step-by-step explanation:
- Existence is a condition that establish that every element of te domain set must be related with some element in the range. Example: if the domain of the function is formed by the elements (1,2,3), and the range is formed by the elements (10,11), the condition is not respected if the element "3" for example, is not linked with 10 or 11 (the two elements of the range set).
- Unicity is a condition that establish that each element of the domain of a relation must be related with <u>only one</u> element of the range. Following the previous example, if the element "1" of the domain can be linked to both the elements of the range (10,11), the relation is not a function.
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Answer:
The image will be congruent to the pre-image because n=1
Step-by-step explanation:
The image dimensions are the original dimensions multiplied by the scale factor. If the scale factor is 1, the image is the same size as the original, hence congruent to the pre-image.