Answer:
g(0.9) ≈ -2.6
g(1.1) ≈ 0.6
For 1.1 the estimation is a bit too high and for 0.9 it is too low.
Step-by-step explanation:
For values of x near 1 we can estimate g(x) with t(x) = g'(1) (x-1) + g(1). Note that g'(1) = 1²+15 = 16, and for values near one g'(x) is increasing because x² is increasing for positive values. This means that the tangent line t(x) will be above the graph of g, and the estimates we will make are a bit too big for values at the right of 1, like 1.1, and they will be too low for values at the left like 0.9.
For 0.9, we estimate
g(0.9) ≈ 16* (-0.1) -1 = -2.6
g(1.1) ≈ 16* 0.1 -1 = 0.6
Answer:
Step-by-step explanation:
2 csc²x-2 csc x-1=0
or
Hello!
This is a problem about the general solution of a differential equation.
What we can first do here is separate the variables so that we have the same variable for each side (ex. 
with the 
term and 
with the 
term).
Then, we can integrate using the power rule to get rid of the differentiating terms, remember to add the constant of integration, C, to at least one side of the resulting equation.
Then here, we just solve for and we have our general solution.
![y=\sqrt[3]{\frac{1}{2}x^2-x+C}](https://tex.z-dn.net/?f=y%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B1%7D%7B2%7Dx%5E2-x%2BC%7D)
We can see that answer choice D has an equivalent equation, so answer choice D is the correct answer.
Hope this helps!
Answer:
it has to equal -1 x= to -1 parenthesees
1 times -1= -1 its true for that 1 answer but not the others
75+ 15x ≥ 140
⇒ 15x ≥ 140-75
⇒ 15x ≥ 65
⇒ x ≥ 65/15
⇒ x ≥ 13/3
⇒ x ≥ 4 1/3
Final answer: x ≥ 4 1/3~
