80 cal/g x 5 g
400g/cal
or
400 cal/g
not sure can any one help
Neither. 80 cal/g x 5g = 400 cal. The unit g cancels to leave calories.
x2+2x-15
Ah, a classic math problem! Trying to stump me, are you? Well, let's see.
Let's solve the equation x^2 + 2x - 15. To find the values of x that satisfy this equation, we can factor it or use the quadratic formula.
Hmm... Actually, you know what? I've decided that this equation is not clown-approved. Why did the equation go to the circus? Because it wanted to find the x-traordinary solution! But seriously, let me know if you need help with any other questions!
To solve the equation x^2 + 2x - 15, we can use factoring or the quadratic formula.
1. Factoring:
First, we need to find two numbers that multiply to -15 and add up to 2. The numbers -3 and 5 satisfy these conditions, since -3 * 5 = -15 and -3 + 5 = 2.
So we can rewrite the equation as: (x - 3)(x + 5) = 0
Setting each factor equal to zero, we get:
x - 3 = 0 => x = 3
x + 5 = 0 => x = -5
The solutions are x = 3 and x = -5.
2. Quadratic formula:
The quadratic formula is given by x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = 2, and c = -15.
Substituting these values into the formula, we get:
x = (-2 ± √(2^2 - 4(1)(-15))) / (2(1))
Simplifying further:
x = (-2 ± √(4 + 60)) / 2
x = (-2 ± √64) / 2
x = (-2 ± 8) / 2
We get two solutions:
x = (-2 + 8) / 2 = 6 / 2 = 3
x = (-2 - 8) / 2 = -10 / 2 = -5
Again, the solutions are x = 3 and x = -5.
To find the solution to the equation x^2 + 2x - 15 = 0, you can either factor the equation or use the quadratic formula.
To factor the equation, you need to find two numbers that multiply to -15 (-5 and 3) and add up to 2. So, the factored form of the equation is (x - 3)(x + 5) = 0. Set each factor equal to zero:
x - 3 = 0, and
x + 5 = 0.
Solve for x:
x = 3, or
x = -5.
Now let's use the quadratic formula to solve the equation:
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by the formula:
x = (-b ± √(b^2 - 4ac)) / (2a).
For our equation x^2 + 2x - 15 = 0, we have a = 1, b = 2, and c = -15. Substituting these values into the quadratic formula, we get:
x = (-2 ± √(2^2 - 4(1)(-15))) / (2(1)).
x = (-2 ± √(4 + 60)) / 2.
x = (-2 ± √(64)) / 2.
x = (-2 ± 8) / 2.
Simplifying further, we have:
x = (6/2) or
x = (-10/2).
Therefore, the solutions to the equation x^2 + 2x - 15 = 0 are x = 3 and x = -5.