You would solve this by determining how many numbers from 1-10 can be multiples of 5. That's obviously 5 and 10. So that's 2/10 or reduced to 1/5. Now the second question you have to find how many numbers from 1-10 are prime. Prime is a number that can only be divided by 1 and itself. So that would leave you with 1, 3, 5, 7. I think where the trick comes into play is if you drew a 5 or 10 the first time, then you didn't place those cards back in the box, you couldn't draw them again. So the answer would 4/10 or 2/5 if you could draw the 5 again. Or I supposed if you had drew the 10 the first time and not the 5. I assume however, they want you to exclude 5 and 10 because it was part of the first section of this equation. So that would mean only 1, 3, and 7 or 3/10.
Answer:
RS = 30 units
Step-by-step explanation:
Given:
Length (PS) = 
Width (PQ) = 
Perimeter = 134 units
Required:
RS (width)
SOLUTION:
Perimeter of a rectangle = 2(length + width)
Plug in the values into the equation
134 = 2[(4y + 1) + (3y + 3)]
134 = 2[4y + 1 + 3y + 3]
134 = 2[7y + 4]
134 = 14y + 8
Subtract 8 from each side of the equation
134 - 8 = 14y
126 = 14y
Divide both sides by 14.
126/14 = 14y/14
9 = y
y = 9
✍️RS = Width (PQ) =
Plug in the value of y
RS = tex] 3(9) + 3 [/tex]
RS = tex] 27 + 3 [/tex]
RS = 30
(f-g)(x) is equivalent to f(x) - g(x)
So that would be (2x²-5) - (x²-4x-8) = 2x² - x² + 4x + 8 - 5 = x² + 4x + 3
Hope this helps.
