If x = 1, then 3*1 = 3 which when modded with 5, we get 3 as a remainder. In other words, 3/5 = 0 remainder 3. We don't use the quotient at all when it comes to modular arithmetic. All we care about is the remainder.
If x = 2, then 3*2 = 6 which leads to remainder 1 when we divide by 5. Therefore, 3x = 1 (mod 5) when x = 2.
If x = 3, then 3*3 = 9 = 4 (mod 5) because 9/5 = 1 remainder 4.
So 3x = 4 (mod 5) when x = 3.
<h3>The final answer is C) 3</h3>
We don't need to check D since x = 3 is a solution and it's smaller than x = 4.
If you wanted to check x = 4, then 3*4 = 12 = 2 (mod 5) because 12/5 yields a remainder of 2.
Answer:

Step-by-step explanation:

Note that √(4 - t²) is defined only as long as 4 - t² ≥ 0, or -2 ≤ t ≤ 2. Then the real integral exists only if -2 ≤ x ≤ 2. (Otherwise we deal with complex numbers.)
If x = 2, then the integral corresponds to the area of a quarter-circle with radius 2. This means that the integral has a maximum value of 1/4 • π • 2² = π.
On the opposite end, if x = -2, then the integral has the same value, but the integral from 0 to -2 is equal to the negative integral from -2 to 0. So the minimum value is -π.
For all x in between, we observe that the integrand is continuous over the rest of its domain, so F(x) is continuous.
Then the range of F(x) is the interval [-π, π].
I am unsure about the very last problem but I can help with the first two
1) (y+1)+4
If we combine the numbers 1 and 4, we get +5 and can isolate the numbers from the variable.
This would give us

2) (6*r)*7
remember that we do not have to explicitly state 6*r
Instead, we can write it as 6r
this helps us get rid of the parentheses
now we can write it as

I hope this helps!:)