Heads, or tails. It's 1/2 a chance, because either way you will either land heads or tails. It would be a probability of 1-(1/2)^5. Can you solve that?
Answer:
Hi, hows life going?
Step-by-step explanation:
Lets get started :)
These questions have asked us to solve by completing the square.
How do we? I have attached a picture, which will explain
6. x² + 2x = 8
→ b is the coefficient of x, which is 2
→ We take half of 2 and square it. Then, we add it to either side
x² + 2x

x² + 2x + 1 = 8 + 1
( x + 1 )( x + 1 ) = 9
( x + 1 )² = 9
x + 1 = + 3 or x + 1 = - 3 x = 2 or x = - 47. x² - 6x = 16
→ We do the same thing we did in the previous question
x² - 6x +
x² - 6x + 9 = 16 + 9
(x - 3)² = 25
x - 3 = + 5 or x - 3 = - 5 x = 8 or x = - 28. x² - 18x = 19
x² - 18x +

x² - 18x + 81 = 19 + 81
( x - 9 )( x - 9 ) = 100
( x - 9 )² = 100

x - 9 = + 10 or x - 9 = -10
x = 19 or x = - 1
9. x² + 3x = 3
x² + 3x +





x +
= +
or x +
x =
or x =
Answer:
The other midpoint is located at coordinates (-9,-2) (Second option)
Step-by-step explanation:
<u>Midpoints</u>
If P(a,b) and Q(c,d) are points in
, the midpoint between them is the point exactly in the center of the line that joins P and Q. Its coordinates are given by


We are given one endpoint at P(1,-2) and the midpoint at M(-4,-2). The other endpoint must be at an equal distance from the midpoint as it is from P. We can see both given points have the same value of y=-2. This simplifies the calculations because we only need to deal with the x-coordinate.
The x-distance from P to M is 1-(-4)=5 units. This means the other endpoint must be 5 units to the left of M:
x (other endpoint)= - 4 - 5 = - 9
So the other midpoint is located at (-9,-2) (Second option)
Answer:
This graph has been translated. Translation is when the image on the graph has been shifted along one or both of the axis. In this case, the graph has been shifted up two units as shown by the operation of addition on the y-value. Remember that translations are rigid movements, so the new image will be congruent to the pre-image.