Answer:
D, g(x) = 1/2 x + 2
Step-by-step explanation:
The second answer IS ONE TO ONE
Answer:
None of the options are correct, but I got g(x)=1/2x+2.
Step-by-step explanation:
You start with the equation. You first need to change f(x) to y, and switch the x and y values. After this, your equation would be x=2y-4. Then, you need to icolate the y value, and to do that you first cancel out the four by adding it to both sides, and now your equation should be x+4=2y. The second step in icolating the y value is dividing everything in the equation by two. Now that the y is icolated, you equation should be y=1/2x+2. You then change the y to be g(x), and that's how I got my answer, g(x)=1/2x+2. You should ask your teacher about this, becuase from my calculations, none of the given responses would be correct. Sorry if that doesn't help you at all but yeah
Answer:
6 1/20 or as a decimal 6.05
Answer:
0, 14
Step-by-step explanation:
-2, is 8 away from 6, so you add 8 to 6 to reflect it and get 14
Answer:
45
Step-by-step explanation:
kari had 15 miles 5x3=15
sondra had 60 miles 15x4=60
40-15=45
Answer:
7.5
Step-by-step explanation:
60% = 0.6
18 / 0.6 = 30 (given number)
25% = 0.25
30 * 0.25 = 7.5
Best of Luck!
Answer:
2.7
Step-by-step explanation:
60/100x = 18
x = 10.8
25/100x = ?
25/100 (10.8) = 2.7
HELP Please
Answer:
it is 10 because 5x2= 10and20-10=10
b. Let h (t) = ( sin (t) + cos (t)) t 4. Find h ′(t).
c. Determine the slope of the tangent line to the curve y = f (x) at the point where a = 1 if f is given by the rule f(x) = e^x sin (x).
d. Find the tangent line approximation L(x) to the function y = g (x) at the point where a = − 1 if g is given by the rule g (x) = ( x^2 + x ) 2^x .
Answer:
A) M'(w) = w^16 * 4^w [ 51 + 3w In4 ]
B) h'(t) = [ cos (t) - sin (t) ] t^4 + [ sin(t) + cos (t) ] 4t^3
C) f'(1) = e' [sin(1) + cos(1) ]
D) g'(a) = 0 - 1/2
L(x) = - 1/2 ( x + 1 )
Step-by-step explanation:
Attached below is the detailed solution of the problem
A) m(w) = 3w^17 * 4^w
M'(w) = w^16 * 4^w [ 51 + 3w In4 ]
B) h(t) = [sin(t) + cos(t) ] t^4
h'(t) = [ cos (t) - sin (t) ] t^4 + [ sin(t) + cos (t) ] 4t^3
C) f(x) = e^x sin (x). at a = -1
f'(1) = e' [sin(1) + cos(1) ]
D) g (x) = ( x^2 + x ) 2^x .
g'(a) = 0 - 1/2
L(x) = - 1/2 ( x + 1 )