to prove that 6720. Sin 40. Sin 60. 80 equals 3 you'd have to divide 16 by 20 multiply subtract both sides by 3 which is 24x + 6x + 8x + -14 equals 40 40 + 20 equals 16 - -40 equals 80 which equals to 3
Answer:
Step-by-step explanation:
Given equation is,
x² + (p + 1)x = 5 - 2p
x² + (p + 1)x - (5 - 2p) = 0
x² + (p + 1)x + (2p - 5) = 0
Properties for the roots of a quadratic equation,
1). Quadratic equation will have two real roots, discriminant will be greater than zero. [(b² - 4ac) > 0]
2). If the equation has exactly one root, discriminant will be zero [(b² - 4ac) = 0]
3). If equation has imaginary roots, discriminant will be less than zero [(b² - 4ac) < 0].
Discriminant of the given equation = 
For real roots,
p² + 2p + 1 - 8p + 20 > 0
p² - 6p + 21 > 0
For all real values of 'p', given equation will be greater than zero.
Answer:
36 pencils
Step-by-step explanation:
Let h and p represent the number of highlighters and the number of pencils, respectively.
Then h + p = 45, and h = 45 - p.
Tom paid a total of $30 for these supplies, with ($2/highligher)(h) + ($0.333/pencil) adding up to that amount.
substituting 45 - p for h in 2h + 0.333p = 30, we get:
2(45 - p) + 0.333p = 30, or
90 - 2p + 0.333p = 30
Combine the constants: 60 = 2p - 0.333p, or 60 = 1.667p
Then p = 60/1.667 = 35.9928, or 36.
Tom bought 36 pencils for $12, and 45-36, or 9, highlighters for $18, for a total purchase of $30. This shows that these calculations are correct.
The domain of a function is where we have a value for x.
Since that's the case the domain of f(x) = {x e R / 1 ≤ x < 5}
We see that we have a value for x = 1 cuz we have a filled circle, but we don't have a value for x = 5, look at the unfilled circle
So, our x can vary between 1 and 5, but can't be 5.
