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lyudmila [28]
3 years ago
10

1.The Student Council is planning a dance. The cost to throw the dance will be $100 for decorations and

Mathematics
2 answers:
Kamila [148]3 years ago
3 0

Answer:

1a)  C = 600 + 5n

1b)  R = 6n

Step-by-step explanation:

Ugo [173]3 years ago
3 0

Answer:

1.

    a) $600+5n = C

    b) 6n = R

Step-by-step explanation:

For the cost to throw the event, we have to add $100+$500, the decorations and the DJ and then we have to add the variable for people times 5 since it is 5 per person. $100+500 = $600 and 5* n (the variable for people in the question)

We are adding the two terms so the final equation will be equal to:

600+5n

for b), we have to use the variable n for tickets times the constant 6 since it costs 6 dollars per person to enter, and that will be equal to the revenue the student council earns or R.

6*n = 6n

Now we have:

6n=R

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THat would be option c  Her cahnge in points would be 4 * -6 = -24.
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Verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial
Mariulka [41]

Answer:

i) Since P(2), P(-1) and P(½) gives 0, then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.

ii) - the sum of the zeros and the corresponding coefficients are the same

-the Sum of the products of roots where 2 are taken at the same time is same as the corresponding coefficient.

-the product of the zeros of the polynomial is same as the corresponding coefficient

Step-by-step explanation:

We are given the cubic polynomial;

p(x) = 2x³ - 3x² - 3x + 2

For us to verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial, we will plug them into the equation and they must give a value of zero.

Thus;

P(2) = 2(2)³ - 3(2)² - 3(2) + 2 = 16 - 12 - 6 + 2 = 0

P(-1) = 2(-1)³ - 3(-1)² - 3(-1) + 2 = -2 - 3 + 3 + 2 = 0

P(½) = 2(½)³ - 3(½)² - 3(½) + 2 = ¼ - ¾ - 3/2 + 2 = -½ + ½ = 0

Since, P(2), P(-1) and P(½) gives 0,then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.

Now, let's verify the relationship between the zeros and the coefficients.

Let the zeros be as follows;

α = 2

β = -1

γ = ½

The coefficients are;

a = 2

b = -3

c = -3

d = 2

So, the relationships are;

α + β + γ = -b/a

αβ + βγ + γα = c/a

αβγ = -d/a

Thus,

First relationship α + β + γ = -b/a gives;

2 - 1 + ½ = -(-3/2)

1½ = 3/2

3/2 = 3/2

LHS = RHS; So, the sum of the zeros and the coefficients are the same

For the second relationship, αβ + βγ + γα = c/a it gives;

2(-1) + (-1)(½) + (½)(2) = -3/2

-2 - 1½ + 1 = -3/2

-1½ - 1½ = -3/2

-3/2 = - 3/2

LHS = RHS, so the Sum of the products of roots where 2 are taken at the same time is same as the coefficient

For the third relationship, αβγ = -d/a gives;

2 * -1 * ½ = -2/2

-1 = - 1

LHS = RHS, so the product of the zeros(roots) is same as the corresponding coefficient

7 0
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5 (-2) + 4 =
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5 (5) + 4 =
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Answer:

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