For this case we must indicate which of the equations shown can be solved using the quadratic formula.
By definition, the quadratic formula is applied to equations of the second degree, of the form:
Option A:
Rewriting we have:
This equation can be solved using the quadratic formula
Option B:
Rewriting we have:
It can not be solved with the quadratic formula.
Option C:
Rewriting we have:
This equation can be solved using the quadratic formula
Option D:
Rewriting we have:
It can not be solved with the quadratic formula.
Answer:
A and C
The first term of the sequence is already given to be 3. Use this value to obtain the second term.
a2 = 2(a1)^2 = 2(3)² = 18
Use the value of the second term to get the third term through the equation,
a3 = 2(a2)² = 2(18)² = 648
Thus, the answer to this item is letter B.
Answer:
the area of the hexagon is 68 cm²
Step-by-step explanation:
Let us observe the image a little, we can find the area of the hexagon equals to the area of the large rectangle minus 4 small triangles surrounding the hexagon. The area of the rectangle is 8 × 12 is 96 cm². The area of 1 small triangle is 4 × 3.5 ÷ 2 = 7, there are 4 triangles, so the total area is 4 × 7 = 28. So the hexagon's area is 96 - 28 = 68 cm²
Hope this will help❤️
Answer:
P(Y = 0) = 0.09
P(Y = 1) = 0.4
P(Y = 2) = 0.32
P(Y = 3) = 0.19
Step-by-step explanation:
Let the events be:
Wednesday
Thursday
Friday
Saturday
Their corresponding probabilities are
Since 
= number of days beyond Wednesday that it takes for both magazines to arrive(so possible values are 0, 1, 2 or 3)
The possible number of outcomes are therefore 
The values associated for each of the outcomes are as follows:
The probability mass function of is,
![P(Y = 0) = 0.3(0.3) = 0.09\\P(Y = 1) = P[(W, T) or (T, W) or (T, T)]\\= [0.3(0.4) + 0.3(0.4) + 0.4(0.4)]\\= 0.4\\\\P(Y = 2) = P[(W, F) or (T, F) or (F, W) or (F, T) or (F, F)]\\= [0.3(0.2) + 0.4(0.2) + 0.2(0.3) + 0.2(0.4) + 0.2(0.2)]\\= 0.32\\\\P(Y = 3) = P[(W, S) or (T, S) or (F, S) or (S, W) or (S, T) or (S, F) or (S, S)]\\= [0.3(0.1) + 0.4(0.1) + 0.2(0.1) + 0.1(0.3) 0.1(0.4) + 0.1(0.2) + 0.1(0.1)]\\= 0.19](https://tex.z-dn.net/?f=P%28Y%20%3D%200%29%20%3D%200.3%280.3%29%20%3D%200.09%5C%5CP%28Y%20%3D%201%29%20%3D%20P%5B%28W%2C%20T%29%20or%20%28T%2C%20W%29%20or%20%28T%2C%20T%29%5D%5C%5C%3D%20%5B0.3%280.4%29%20%2B%200.3%280.4%29%20%2B%200.4%280.4%29%5D%5C%5C%3D%200.4%5C%5C%5C%5CP%28Y%20%3D%202%29%20%3D%20P%5B%28W%2C%20F%29%20or%20%28T%2C%20F%29%20or%20%28F%2C%20W%29%20or%20%28F%2C%20T%29%20or%20%28F%2C%20F%29%5D%5C%5C%3D%20%5B0.3%280.2%29%20%2B%200.4%280.2%29%20%2B%200.2%280.3%29%20%2B%200.2%280.4%29%20%2B%200.2%280.2%29%5D%5C%5C%3D%200.32%5C%5C%5C%5CP%28Y%20%3D%203%29%20%3D%20P%5B%28W%2C%20S%29%20or%20%28T%2C%20S%29%20or%20%28F%2C%20S%29%20or%20%28S%2C%20W%29%20or%20%28S%2C%20T%29%20or%20%28S%2C%20F%29%20or%20%28S%2C%20S%29%5D%5C%5C%3D%20%5B0.3%280.1%29%20%2B%200.4%280.1%29%20%2B%200.2%280.1%29%20%2B%200.1%280.3%29%200.1%280.4%29%20%2B%200.1%280.2%29%20%2B%200.1%280.1%29%5D%5C%5C%3D%200.19)
