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Artyom0805 [142]
3 years ago
10

GIVING BRANLIEST!!!

Mathematics
1 answer:
lidiya [134]3 years ago
5 0

Answer:

I believe it's (6,10).

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What is the inverse of the function??
Firlakuza [10]
\(\begin{matrix}
y=ln(x+3)\\
e^y=x+3\\
e^y-3=x\\
swap ~x~and~y\\
 \huge\underline{y=e^x-3}}
 \end{matrix}\)


To get the inverse of any function f(x) [ just make sure that it passes the horizontal line test] follow the steps:

- Solve for x, i.e separate x in one side and the other terms in the other side, included f(x).

- Swap x and f(x).

- The result is the inverse of the original function f^{-1}(x).

-------------------------

Apply that on your function
\(\begin{matrix}
y=ln(x+3)\\
e^y=x+3\\
e^y-3=x\\
swap ~x~and~y\\
 \huge\underline{y=e^x-3}}
 \end{matrix}\)




5 0
3 years ago
Translate the phrase into an algebraic expression. <br><br> the sum of b and 4
weqwewe [10]

Answer:

b+4

hope this helps!

4 0
3 years ago
Which equation is correct? 2x+38) - 5x +4 A 7x - 41 B 7x + 41 C 36x - 2<br>or just write your own
bija089 [108]

Answer:

-3x+42

Step-by-step explanation:

(2x+38) - 5x +4

2x+38 - 5x +4

2x+38+ - 5x +4

2x+-5x+38+4

-3x+42

5 0
4 years ago
Integrate t sec^2 (2t) dt
Neporo4naja [7]
F = t ⇨ df = dt 
dg = sec² 2t dt ⇨ g = (1/2) tan 2t 
⇔ 
integral of t sec² 2t dt = (1/2) t tan 2t - (1/2) integral of tan 2t dt 
u = 2t ⇨ du = 2 dt 

As integral of tan u = - ln (cos (u)), you get : 

integral of t sec² 2t dt = (1/4) ln (cos (u)) + (1/2) t tan 2t + constant 
integral of t sec² 2t dt = (1/2) t tan 2t + (1/4) ln (cos (2t)) + constant 
integral of t sec² 2t dt = (1/4) (2t tan 2t + ln (cos (2t))) + constant ⇦ answer 
8 0
4 years ago
Read 2 more answers
I NEED HELP ITS FOR A GRADE THIS IS LAST QUESTION I GIVE U BRAINLIEST IF U GET IT RIGHT
yulyashka [42]

Answer:

1rst is 2

2nd is 28 4/5

3rd is 26 4/5

3 0
3 years ago
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