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vazorg [7]
2 years ago
7

Mike bought 7 new baseball trading cards to add to his collection. His dog ate half of his collection. There are now only 25 car

ds left. How many cards did Mike have to start with?
Mathematics
1 answer:
RSB [31]2 years ago
6 0

Answer:50

Step-by-step explanation:

Bc 25+25 is 50.

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Michael says their homes are at opposite elevations because they are in opposite directions. Is he correct? Explain.
ratelena [41]
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3 0
2 years ago
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You start out with $20 and then spend money in a store where every item is $3. Y = -3x + 20.
hodyreva [135]

D

Step-by-step explanation:

3 0
3 years ago
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If you are allowed to use numbers 1 – 20 and need to choose the passcode of an exact 4 digit code, how many possibilities are th
VMariaS [17]

Since in a pass code, the placement of the digits is important, therefore this means that to solve for the total number of possibilities we have to make use of the principle of Permutation. The formula for calculating the total number of possibilities using Permutation is given as:

P = n! / (n – r)!

where,

n = is the total amount of numbers to choose from = 20

r = is the total number of digits needed in the passcode = 4

 

Therefore solving for the total possibilities P:

P = 20! / (20 – 4)!

P = 20! / 16!

P = 116,280

 

<span>Hence there are a total of 116,280 possibilities of pass codes.</span>

4 0
2 years ago
Neil has 3 partially full cans of white paint. They contain 1/3 gallon, 1/5 gallon and 1/2 gallon of paint. About how much paint
lorasvet [3.4K]

Answer:

31/30 or 1 1/30

Step-by-step explanation:

1/3+1/5+1/2

6 0
2 years ago
Check answer please
Cerrena [4.2K]
The fourth or the D) Option is correct.

To find the new induced matrix via a scalar quantified multiplication we have to multiply the scalar quantity with each element surrounded and provided in a composed (In this case) 3×3 or three times three matrix comprising 3 columns and 3 rows for each element which is having a valued numerical in each and every position.

Multiply the scalar quantity with each element with respect to its row and column positioning that is,

Row × Column. So;

(1 × 1) × 7, (2 × 1) × 7, (3 × 1) × 7, (1 × 2) × 7, (2 × 2) × 7, (3 × 2) × 7, (1 × 3) × 7, (2 × 3) × 7 and (3 × 3) × 7. This will provide the final answer, that is, the D) Option.

To interpret and make it more interesting in LaTeX form. Here is the solution with LaTeX induced matrix.

\mathcal{A = \begin{bmatrix}1 & 0 & 3 \\ 2 & -1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}

\mathbf{\therefore \quad 7A = 7 \times \begin{bmatrix}1 & 0 & 3 \\ 2 & - 1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}

\mathbf{\therefore \quad \begin{bmatrix}7 \times 1 & 7 \times 0 & 7 \times 3 \\ 7 \times 2 & 7 \times -1 & 7 \times 2 \\ 7 \times 0 & 7 \times 2 & 7 \times 1 \\ \end{bmatrix}}

\therefore \quad \begin{\bmatrix}7 & 14 & 0 \\ 0 & -7 & 14 \\ 21 & 14 & 7 \end{bmatrix}

Hope it helps.
5 0
3 years ago
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