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3 years ago

Which rule represents the translation from the pre-image, ΔABC, to the image, ΔA'B'C'?

Mathematics

2 answers:

V125BC [204]3 years ago

8 0

Answer:

(x, y) → (x + 7, y – 6).

Step-by-step explanation:

ANEK [815]3 years ago

6 0

From the figure, it can be seen that the mage A'B'C' is obtained from the pre-image ABC by translating/shifting the vertices of the image by 7 units to the right and 6 units down.

Therefore, the rule represents the translation from the pre-image, ΔABC, to the image, ΔA'B'C' is (x, y) → (x + 7, y – 6).

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3 years ago

(cotx+cscx)/(sinx+tanx)

Butoxors [25]

Answer: $\bold{\dfrac{cot(x)}{sin(x)}}$

Step-by-step explanation:

Convert everything to "sin" and "cos" and then cancel out the common factors.

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$\text{Simplify:}\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)+sin(x)}{cos(x)}\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)cos(x)+sin(x)}\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)(cos(x)+1)}\bigg)$

$\text{Cross out the common factor of (cos(x) + 1) from the top and bottom}:\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)}\bigg)\\\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times cot(x)}\qquad \rightarrow \qquad \dfrac{cot(x)}{sin(x)}$

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3 years ago