For how many real values of x is an integer?

A good place to start is to set to y. That would mean we are looking for to be an integer. Clearly, , because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since is a radical, it only outputs values from , which means y is on the closed interval: .

With that, we don't really have to consider y anymore, since we know the interval that is on.

Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval , which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:

Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.

at least two

step-by-step explanation: