Use the IVT to prove that one solution to is between x=1 and x=2. Please show work and if your answer makes no sense I will report. Thanks

Let f(x) = x³ + x - 5. f(x) is a polynomial so it's continuous everywhere on its domain (all real numbers). Since

f (1) = 1³ + 1 - 5 = -3 < 0

and

f (2) = 2³ + 2 - 5 = 5 > 0

it follows by the intermediate value theorem that there at least one number x = c between 1 and 2 for which f(c) = 0.

astronomers is your final answer! : )

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions