Let g be a differentiable function with g(-1)=5 and g'(-1)=2. What is the value of the approximation of g(-0.9) using the function's local linear approximation at x= -1? Options: 5.1, 5.2, 5.3, or 5.4

5.2

Step-by-step explanation:

g(x) = mx +b

g'(x) = m = 2 for all x

g(-1) = 2(-1) + b = 5

-2 + b = 5

b = 7

So, g(x) = 2x + 7

Therefore g(-.9) = 2(-.9) + 7

= -1.8 + 7 = 5.2

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.

i need more details sorry

step-by-step explanation: