If f(x)=x2+10sinx, show that there is a number c such that f(c)=1000.

F(x) is continuous for all x.

Pick a point and show that f(x) is either negative or positive. Pick another point and show that f(x) is negative, if positive, or positive, if negative.

At x = 30, f(30) - 1000 = 900 + 10sin(30) - 1000 ≤ 0
Now, show at another point f(x) - 1000 is positive, and hence, there would be root between 30 and such point.

Let's pick 40.
At x = 40, f(40) - 1000 = 1600 + 10sin(40) - 1000 ≥ 0

Since f(x) - 1000 is continuous, there lies a root between 30 and 40, and hence, 30 ≤ c ≤ 40

Hi! mark me brainliest if i : )y = 0.213x – 352.0    represents this linear model shown in the data table.plug in the values of x into the equation for a double check.let's try 1980.y = 0.213(1980) – 352.0y = 69.74which is closest to the 70.1 whereas other options do not satisfy the condition.