Answer:
![x = 5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}](https://tex.z-dn.net/?f=x%20%3D%205%2B%5Csqrt%7B31%7D%5C%2C%5C%2C%20and%5C%2C%5C%2C%20x%3D5-%5Csqrt%7B31%7D)
Step-by-step explanation:
We need to solve the quadratic equation
6 = x^2 -10x
Rearranging we get,
x^2-10x-6=0
Using quadratic formula to solve the quadratic equation
![x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
a= 1, b =-10 and c=6
Putting values in the quadratic formula
![x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(-6)}}{2(1)}\\x=\frac{10\pm\sqrt{100+24}}{2}\\x=\frac{10\pm\sqrt{124}}{2}\\x=\frac{10\pm\sqrt{2*2*31}}{2}\\x=\frac{10\pm\sqrt{2^2*31}}{2}\\x=\frac{10\pm2\sqrt{31}}{2}\\x = 5\pm\sqrt{31}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%28-10%29%5Cpm%5Csqrt%7B%28-10%29%5E2-4%281%29%28-6%29%7D%7D%7B2%281%29%7D%5C%5Cx%3D%5Cfrac%7B10%5Cpm%5Csqrt%7B100%2B24%7D%7D%7B2%7D%5C%5Cx%3D%5Cfrac%7B10%5Cpm%5Csqrt%7B124%7D%7D%7B2%7D%5C%5Cx%3D%5Cfrac%7B10%5Cpm%5Csqrt%7B2%2A2%2A31%7D%7D%7B2%7D%5C%5Cx%3D%5Cfrac%7B10%5Cpm%5Csqrt%7B2%5E2%2A31%7D%7D%7B2%7D%5C%5Cx%3D%5Cfrac%7B10%5Cpm2%5Csqrt%7B31%7D%7D%7B2%7D%5C%5Cx%20%3D%205%5Cpm%5Csqrt%7B31%7D)
So, ![x=5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}](https://tex.z-dn.net/?f=x%3D5%2B%5Csqrt%7B31%7D%5C%2C%5C%2C%20and%5C%2C%5C%2C%20x%3D5-%5Csqrt%7B31%7D)
First of all, your fractions that you wrote were wrong. You flipped the numbers over. When you're finding the slope of a graph, the number on the y-axis is always the numerator while the x-axis is the denominator.
Your first answer would be the actual slope- which is 3.
Your second answer would be the number of products that company would sell if the got 1,000 web views So just multiply 3 by 1,000. That means that the company will sell 3,000 products.
Answer:
The time taken for the projectile to reach the given height is 1.2 s.
Step-by-step explanation:
Given;
height of travel, h = 25 ft
initial velocity of the projectile, u = 15 ft/s
The time taken for the projectile to travel a height of 25 ft is given by the following kinematic equation;
h = ut + ¹/₂gt²
25 = 15t + ¹/₂(9.8)t²
25 = 15t + 4.9t²
4.9t² + 15t - 25 = 0
solving the quadratic equation, we will have the following solution of t;
t = 1.2 s
Therefore, the time taken for the projectile to reach the given height is 1.2 s.
I: 635=dimes*10+quarters*25
or shorter
635=d*10+q*25
II: d=q*3+3
Substitute II (d=3q+3) into I to replace d:
635=d*10+q*25
635=(3q+3)*10+q*25
635=30+30q+25q
605=55q
11=q
-> 11 quarters
"What expression represents the number of dimes if q represent quarters"
insert q=11 into II:
II: d=q*3+3
d=11*3+3
d=33+3
d=36