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RideAnS [48]
3 years ago
8

An image is 2 units to the left of the pre-image.

Mathematics
2 answers:
kap26 [50]3 years ago
7 0

Answer:

Santana’s reasoning is not correct. The translation should be 2 units to the left, not to the right.

Step-by-step explanation:

I got it right.

stepladder [879]3 years ago
5 0

Answer:

Santana's thinking is not correct, because the correct translation is 2 units to the left, not right.

Step-by-step explanation:

I just got that question and I got it right.

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PLEASE ANSWER ASASP FOR BRAINLEST!!!!!!!!!!!!!
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7 0
3 years ago
The range of y = Arcsin x is<br> kle<br> A. True<br> B. False
Yuri [45]

Answer:

I guess is

Step-by-step explanation:

False I guess

6 0
3 years ago
If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=
Alexxx [7]
\bf f(x)=y=2x+sin(x)&#10;\\\\\\&#10;inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)&#10;\\\\\\&#10;\textit{now, the "y" in the inverse, is really just g(x)}&#10;\\\\\\&#10;\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\&#10;-----------------------------\\\\

\bf \textit{let's use implicit differentiation}\\\\&#10;1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}&#10;\\\\\\&#10;1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\&#10;-----------------------------\\\\&#10;g'(2)=\cfrac{1}{2+cos[g(2)]}

now, if we just knew what g(2)  is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that  means if we use that value in f(x),   f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus    2 = 2x+sin(x)

\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2&#10;\\\\\\&#10;-----------------------------\\\\&#10;g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
5 0
2 years ago
Which expression is the factorization of x2 + 10x + 21?
maxonik [38]

Answer:

(x + 3)(x + 7)

Step-by-step explanation:

Find two numbers that when added up to , they ALSO have to multiply up to 21. This is simple because of the fact that there is no leading coefficient greater than 1⃣.

7 0
2 years ago
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