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ANTONII [103]
2 years ago
11

Aden buys 15 collectible coins per month. Caliana sells 5 coins from her collection of 100 each month. When will Aden have more

coins than Caliana
Mathematics
2 answers:
Ratling [72]2 years ago
6 0

Answer:

Answer: putting it into a chart, when Calinana has 75, Aden will have 90

Step-by-step explanation:

Step-by-step explanation:

user100 [1]2 years ago
5 0

Answer: putting it into a chart, when Calinana has 75, Aden will have 90

Step-by-step explanation:

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I believe it’s 80 hope it helps
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How many square feet of cedar are needed to line the four walls of a closet that is a square prism 5 1/2 ft on each edge
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22 square feet because there is 5 1/2 on each wall
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Convert to slope form: 3x-4y=8 then graph the line on the grid.​
alexandr402 [8]

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Step-by-step explanation:

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6 0
3 years ago
1. Simplify the expression and show each step.
Tema [17]

Answer:

1. 3x + 2

2. 0.25x - 6.5

3. $5.50 per sandwich

Step-by-step explanation:

1.

6x + 1 - (3x - 1)

PEMDAS says solve everything in the parentheses first

There is nothing left to simplify within the parentheses.

Since there's also no exponents, multiplication, or division, let's combine like terms starting from left to right

6x - 3x + 1 + 1

Negative + Negative = Positive, hence why it became 1 + 1

3x + 2

2.

(3x - 1) - 2.75 (x + 2)

PEMDAS says to do multiplication next (since parentheses are fully simplified and there are no exponents)

(3x - 1) - 2.75x - 5.5

Negative + Positive = Negative, hence why it became -5.5

3x - 1 - 2.75x - 5.5

Combine like terms

3x - 2.75x - 1 - 5.5

0.25x - 6.5

3.

Set up an equation:

Variable x = cost of sandwiches

4x + 4(2) = 30

4x + 8 = 30

Subtract 8 from both sides to isolate the variable

4x + 8 - 8 = 30 - 8

4x = 22

Divide both sides by 4

4x ÷ 4 = 22 ÷ 4

x = 5.5

$5.50 per sandwich

4 0
3 years ago
Solve the system of equations using matrices. Use Gaussian elimination with back-substitution.
VikaD [51]

In augmented matrix form, the system is equivalent to

\begin{bmatrix}1&1&1&-5\\1&-1&3&-1\\4&1&1&-2\end{bmatrix}

Subtract row 1 from row 2, and 4(row 1) from row 3:

\begin{bmatrix}1&1&1&-5\\0&-2&2&4\\0&-3&-3&18\end{bmatrix}

Divide through row 2 by -2, and through row 3 by -3:

\begin{bmatrix}1&1&1&-5\\0&1&-1&-2\\0&1&1&-6\end{bmatrix}

Subtract row 2 from row 3:

\begin{bmatrix}1&1&1&-5\\0&1&-1&-2\\0&0&2&-4\end{bmatrix}

Divide through row 3 by 2:

\begin{bmatrix}1&1&1&-5\\0&1&-1&-2\\0&0&1&-2\end{bmatrix}

The last row tells you \boxed{z=-2}. Then

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and

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4 0
3 years ago
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