Answer and Step-by-step explanation:
If the radius is 5, then we can plug 5 into the area of a circle equation.
A = 
A = 
A = 25
<u>The area of a circle with radius 5 is 25</u><u>.</u>
<em><u>#teamtrees #PAW (Plant And Water)</u></em>
Let's represent the two numbers by x and y. Then xy=60. The smaller number here is x=y-7.
Then (y-7)y=60, or y^2 - 7y - 60 = 0. Use the quadratic formula to (1) determine whether y has real values and (2) to determine those values if they are real:
discriminant = b^2 - 4ac; here the discriminant is (-7)^2 - 4(1)(-60) = 191. Because the discriminant is positive, this equation has two real, unequal roots, which are
-(-7) + sqrt(191)
y = -------------------------
-2(1)
and
-(-7) - sqrt(191)
y = ------------------------- = 3.41 (approximately)
-2(1)
Unfortunately, this doesn't make sense, since the LCM of two numbers is generally an integer.
Try thinking this way: If the LCM is 60, then xy = 60. What would happen if x=5 and y=12? Is xy = 60? Yes. Is 5 seven less than 12? Yes.
The answer I believe is c
Answer:
X = 4
Y = 30
i think.
Step-by-step explanation:
