K is a prime number and f(x) = k has no solutions.
Find the possible values of k.​

Suppose x∈R is such that x+20x=p is a prime. Then x2−px+20=0 , and this equation has a solution x∈R if and only if p2–80≥0 . So the primes which admit no real solution are those that satisfy p<80−−√ , or 2 , 3 , 5 , and 7 . ■

Step-by-step explanation:

E? <3

no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3

the answer will be