Answer:
3. 3(x+y)(x-y)
4. (y²+4)(y+2)(y-2)
5. 2(3x+1)(x+2)
Step-by-Step:
3. 3(x²-y²) = 3(x+y)(x-y)
4. y⁴-16 = (y²+4)(y²-4)= (y²+4)(y+2)(y-2)
5. 6x²+14x+4 = 2(3x²+7x+2) = 2(3x+1)(x+2)
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(I have verified my answers on an online calculator)
Answer:
Option C is right
C. They are independent because, based on the probability, the first ace was replaced before drawing the second ace.
Step-by-step explanation:
Given that the probability of drawing two aces from a standard deck is 0.0059
If first card is drawn and replaced then this probability would change. By making draws with replacement we make each event independent of the other
Drawing ace in I draw has probability equal to 4/52, when we replace the I card again drawing age has probability equal to same 4/52
So if the two draws are defined as event A and event B, the events are independent
C. They are independent because, based on the probability, the first ace was replaced before drawing the second ace.
The base is a square, therefore all the sides of the base are the same. This also means all the triangles have the same area.
10*10 = 100 cm^2 for the base
0.5*10*12 = 60 cm^2 for each triangle
100 + 4(60) = 340 cm^2
15 = x% from 200
<span>15 = </span><span> * 200</span>
15 = 2x / ÷ 2 (both sides)
<u>x = 7.5 [percent]</u>
Answer:
t= <u>r-21</u>
7
Step-by-step explanation:
r = 7(t+3)
expanding the bracket
r = 7t + 21
bringing t to the left hand side
7t = r - 21
divide both sides by seven
t= <u>r-21</u>
7