5n+9=2n
5n+9−2n=2n−2n
3n+9=0
3n+9−9=0−9
3n=−9
3n/3=−9/3
n=−3
There's a special case of the quadratic formula called the Shakespeare Quadratic Formula (2b or -2b) that applies when the linear coefficient is even.
![x^2-2Bx + C \textrm{ has zeros }x = B \pm\sqrt{B^2 - C}](https://tex.z-dn.net/?f=x%5E2-2Bx%20%2B%20C%20%5Ctextrm%7B%20has%20zeros%20%7Dx%20%3D%20B%20%5Cpm%5Csqrt%7BB%5E2%20-%20C%7D)
If you're planning on solving a lot of quadratic equations quickly, it's good to know.
Here we have
![x^2 + 4x - 7 = 0](https://tex.z-dn.net/?f=x%5E2%20%2B%204x%20-%207%20%3D%200)
so
![x = -2 \pm \sqrt{2^2 - -7}](https://tex.z-dn.net/?f=x%20%3D%20-2%20%5Cpm%20%5Csqrt%7B2%5E2%20-%20-7%7D)
![x = -2 \pm \sqrt{11}](https://tex.z-dn.net/?f=x%20%3D%20-2%20%5Cpm%20%5Csqrt%7B11%7D)
Answer/Step-by-step explanation:
7x- 4 = -182
<em>Step:</em>
Add 4 to both sides
7x - 4 + 4 = -182 + 4
Simplify
7x = -178
Divide both sides by 7
7x/7 = -178/7
Simplify
x = -178/7
Decimal form = -25.4285714286
<u><em>~lenvy~</em></u>
Answer:
P(C4) = 0.0711
Step-by-step explanation:
consider the first draw = 15/23 since it cannot be a blue ball
The second draw = 21/29 since 6 more red balls will be added after the draw since a blue ball cannot be drawn
the third draw = 27/35 since 6 more red balls will be added after each draw since a blue ball cannot be drawn
therefore the total number of red balls will be = 15 + 6 + 6 + 6 = 33 red balls after the 4th draw. the total ball now in the urn= 33 red + 4 blue = 41
Hence the probability of drawing a blue ball at the fourth draw after drawing red balls at the previous attempts = 8/41
P(C4) = P ( fourth ball is blue ) * P( first ball red)*P(second ball red) *P(third ball red )
= (8/41) * (15/23) * (21/29)* (27/35) = 0.0711