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Setler [38]
2 years ago
9

Jermaine, Keira, and Leon are buying supplies for an art class. Jermaine bought 2 canvases, 4 tubes of paint, and 2 paint brushe

s for $34. Keira spent $22 on 1 canvas, 3 tubes of paint, and 1 paint brush. Leon already has plenty of paint, so he bought 3 canvases and 2 paint brushes for $17. How much did each type of material cost?​
Mathematics
1 answer:
maria [59]2 years ago
7 0

Answer:

canvas $3

tube of paint $5

paint brush $4

Step-by-step explanation:

We need to choose variables for the prices of the different items, translate the sentences into equations, and solve a system of equations.

1. Define variables

Let c = price of 1 canvas, t = price of 1 tube of paint, and b = price of 1 brush.

2. Translate sentences into equations

"Jermaine bought 2 canvases, 4 tubes of paint, and 2 paint brushes for $34."

2c + 4t + 2b = 34

"Keira spent $22 on 1 canvas, 3 tubes of paint, and 1 paint brush."

c + 3t + b = 22

"Leon already has plenty of paint, so he bought 3 canvases and 2 paint brushes for $17."

3c + 2b = 17

The system of equations is

2c + 4t + 2b = 34       Eq. 1

c + 3t + b = 22           Eq. 2

3c + 2b = 17               Eq. 3

Since the third equation has only the variables c and b, we use the first equations to eliminate variable t and give us an equation in only c and b.

3 * Eq. 1 - 4 * Eq. 2

2c + 2b = 14

c + b = 7            Eq. 4

Eq. 3 and Eq. 4 form a system of equations in two variables.

3c + 2b = 17      Eq. 3

c + b = 7          Eq. 4

Solve Eq. 4 for b.

b = 7 - c        Eq. 5

Substitute into Eq. 3.

3c + 2(7 - c) = 17

3c + 14 - 2c = 17

c = 3

Plug in c = 3 into Eq. 5.

b = 7 - 3

b = 4

Plug in c = 3 and b = 3 into Eq. 2.

c + 3t + b = 22           Eq. 2

3 + 3t + 4 = 22

3t + 7 = 22

3t = 15

t = 5

Answer:

canvas $3

tube of paint $5

paint brush $4

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Answer:

Step-by-step explanation:

The area of a regular polygon is

A=\frac{1}{2}ap where a is the apothem and p is the perimeter. Perimeter is simple: just count the number of sides the polygon has and multiply it by 3 (the length of one side).

p = 3(9) is 27 inches for the perimeter.

We have what we need now to find the area:

A=\frac{1}{2}(4.1)(27) and

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5 0
2 years ago
What is 5^4 equal please help me.
Nina [5.8K]
5^4 equals 625 your welcome
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3 years ago
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Step-by-step explanation:

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2 years ago
Find three positive consecutive integers such that the product of the smallest and the largest is 17 more than three times the m
trapecia [35]
There are two answers to this question
the first possible answer is 5, 6, and 7 and the second possible answer is -4, -3, and -2


Step-by-step explanation:

let the three numbers be (x-1), x, and (x+1)
the product of the smallest and largest number is 17 more than 3 times the middle number, or
(x-1)(x+1) = 3x+17, or
x^2-1 = 3+17, or
x^2-3x-18 = 0
(x-6)(x+3) = 0
So x = 6 or -3
Then just subtract one and add one to get the other integers. the three numbers are 5, 6, and 7 or -4, -3, and -2
3 0
2 years ago
Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.
Aleks [24]

Answer:

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

Step-by-step explanation:

The actual Series is::

\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}

The method we are going to use is comparison method:

According to comparison method, we have:

\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}

Now:

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty}  \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

\sum_{n=1}^{inf}\frac{1}{n^p}

if p>1 then series converges. In oue case we have:

\sum_{n=1}^{inf}b_n=\frac{1}{n^4}

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

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3 years ago
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