Answer:
(g+f)(x)=(2^x+x-3)^(1/2)
Step-by-step explanation:
Given
f(x)= 2^(x/2)
And
g(x)= √(x-3)
We have to find (g+f)(x)
In order to find (g+f)(x), both the functions are added and simplified.
So,
(g+f)(x)= √(x-3)+2^(x/2)
The power x/2 can be written as a product of x*(1/2)
(g+f)(x)= √(x-3)+(2)^(1/2*x)
We also know that square root dissolves into power ½
(g+f)(x)=(x-3)^(1/2)+(2)^(1/2*x)
We can see that power ½ is common in both functions so taking it out
(g+f)(x)=(x-3+2^x)^(1/2)
Arranging the terms
(g+f)(x)=(2^x+x-3)^(1/2) ..
If UT is the midsegment of QRS, then SQ = 2(2x + 2) = 4x + 4
Then,
(3x + 1)/(7x - 1) = (2x + 2)/(4x + 4) = 1/2
2(3x + 1) = 7x - 1
6x + 2 = 7x - 1
7x - 6x = 2 + 1
x = 3
Therefore, SQ = 4(3) + 4 = 12 + 4 = 16.
Answering pretty sure yes
Step-by-step explanation:
Answer:
5.5 units
Step-by-step explanation:
I'm not 100% sure if this is right, since it could be 5 or 6, but since there is a slight diagonality to it, i would say 5.5 units.
Answer:
28.0
Step-by-step explanation:
if rounding to the nearest tenth then the number in the hundredth determines what to do - if the number in hundredths is '5 and above, give it a shove'
meaning in this case the 7 bumps the 9 up by one; but a nine will carry over and bump the number in the ones place up by one