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nevsk [136]
3 years ago
5

Mary is planning to give 6 friends a book and some candy. Each book was $4.50. She wants to spend no more than $39.00. How many

pieces of candy can she give to each friend if the candy is $0.25 for each piece?​
Mathematics
1 answer:
VARVARA [1.3K]3 years ago
8 0

Answer:

8

Step-by-step explanation:

First multiply $4.50 and 6, you get $27.

Subtract $27 from $39, you get $12 left.

Divide $12 by .25cents, you get 48 pieces of candy.

Divide 48 by 6, you get 8.

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It is possible to have a triangle with those lengths

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a term will be the expression between the + or - signs

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2 years ago
Divide f(x) by d(x), and write a summary statement in the form indicated.
Mandarinka [93]

Answer:

The correct option is B)  f(x) =(x^2+1)(x^2+4x+5)

Step-by-step explanation:

Consider the provided function.

f(x) = x^4 + 4x^3 + 6x^2 + 4x + 5 and d(x) = x^2+1

We need to divide f(x) by d(x)

As we know: Dividend = Divisor × Quotient + Remainder

In the above function f(x) is dividend and divisor is d(x)

Divide the leading term of the dividend by the leading term of the divisor:\frac{x^4}{x^2}=x^2

Write the calculated result in upper part of the table.

Multiply it by the divisor: x^2(x^2+1)=x^4+x^2

Now Subtract the dividend from the obtained result:

(x^4 + 4x^3 + 6x^2 + 4x + 5)-(x^4-x^2)=4x^3+5x^2+4x+5

Again divide the leading term of the obtained remainder by the leading term of the divisor: \frac{4x^3}{x^2}=4x

Write the calculated result in upper part of the table.

Multiply it by the divisor: 4x(x^2+1)=4x^3+4x

Subtract the dividend:

(4x^3+5x^2+4x+5)-(4x^3+4x)=5x^2+5

Divide the leading term of the obtained remainder by the leading term of the divisor: \frac{5x^2}{x^2}=5

Multiply it by the divisor: 5(x^2+1)=5x^2+5

Subtract the dividend:

(5x^2+5)-(5x^2+5)=0

Therefore,

Dividend = x^4 + 4x^3 + 6x^2 + 4x + 5

Divisor = x^2+1

Quotient = x^2+4x+5

Remainder = 0

Dividend = Divisor × Quotient + Remainder

f(x) = (x^2+1)(x^2+4x+5)

Hence, the correct option is B)  f(x) =(x^2+1)(x^2+4x+5)

3 0
3 years ago
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