Answer:
120th caller
Step-by-step explanation:
According to the problem, calculation of the given data are as follows,
$100 bill = 30th caller
Movie tickets = 40th caller
So, to find out first to win both prizes, we will use LCM method,
LCM of 30 and 40 = 10 | 30 , 40
| 3 , 4
LCM of 30 and 40 = 10 × 3 × 4 = 120
Hence, 120th caller will be the first to win both prizes.
∠GHJ and ∠RST are complementary
⇒ ∠GHJ and ∠RST = 90°
And ∠RST and ∠ABC are supplementary
⇒ ∠RST and ∠ABC = 180°
Also, ∠GHJ = x°
⇒ ∠RST = 90° - x°
⇒ ∠ABC = 180° - ∠RST
⇒ ∠ABC = 180° - (90° - x°)
⇒ ∠ABC = 90° + x°
Answer:
We have the function:
r = -3 + 4*cos(θ)
And we want to find the value of θ where we have the maximum value of r.
For this, we can see that the cosine function has a positive coeficient, so when the cosine function has a maximum, also does the value of r.
We know that the meaximum value of the cosine is 1, and it happens when:
θ = 2n*pi, for any integer value of n.
Then the answer is θ = 2n*pi, in this point we have:
r = -3 + 4*cos (2n*pi) = -3 + 4 = 1
The maximum value of r is 7
(while you may have a biger, in magnitude, value for r if you select the negative solutions for the cosine, you need to remember that the radius must be always a positive number)
For the first question, the chance of rolling a six is a 1/6 chance because there is one six and six different sides that could be rolled. As a percent, it would be 16.6 %. The probability of rolling an odd number on the second roll would be a 3/6 chance, which as a percent is 50 %. For the second question, both probabilities are 33 % because in both instances you are drawing three cards from nine total, so it would be 3/9. For the third question, the probability of drawing a blue marble is 3/5, because there are three blue marbles and five total marbles. As a percent, this is 60 %. Following this up with a green marble would be 2/4, because there are now 2 green marbles and four total marbles. One of the marbles was not replaced, so we have one less marble. 2/4 as a percent is 50%.
C. x³-4x²-16x+24.
In order to solve this problem we have to use the product of the polynomials where each monomial of the first polynomial is multiplied by all the monomials that form the second polynomial. Afterwards, the similar monomials are added or subtracted.
Multiply the polynomials (x-6)(x²+2x-4)
Multiply eac monomial of the first polynomial by all the monimials of the second polynomial:
(x)(x²)+x(2x)-(x)(4) - (6)(x²) - (6)(2x) - (6)(-4)
x³+2x²-4x -6x²-12x+24
Ordering the similar monomials:
x³+(2x²-6x²)+(-4x - 12x)+24
Getting as result:
x³-4x²-16x+24
