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pishuonlain [190]
3 years ago
14

A school is buying 35 new computers. If the price of each computer is $1,099,

Mathematics
2 answers:
mote1985 [20]3 years ago
7 0

Answer:

$38,465

Step-by-step explanation:

35 + 1099 = $38,465

dolphi86 [110]3 years ago
6 0

Answer:

$38,465

Step-by-step explanation:

35 x 1,099 = 38,465

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A car rents for $250 per week plus $0.50 per mile.find the rental cost for a two-week trip of 500 miles for a group of three peo
Artyom0805 [142]
So 2 weeks mean
250x 2= 500
Then .50 per mile so 500 miles mean
.50 x 500= 250
In total it will be
500+250= 750
5 0
3 years ago
jody wants to figure out how he spends his time. currently, he spends 12 days out of a 30-day month practicing the piano. How mu
shtirl [24]

Answer:

percent- 40%

decimal- 0.4

percent- 2/5

Step-by-step explanation:

you divide 30 by 12 because he practices piano 12/30 days of the month

3 0
3 years ago
PLease help me out with this...I will mark the first one as the brainiest answer.
Slav-nsk [51]
You cannot add row 2 to column 3 because they have different dimensions. You can do any of the other operations, but the only one that makes any sort of sense is ...
  Multiply row 2 by -1 and add it to row 3

_____
It makes no sense to multiply a row by zero. That makes the entire row zero and makes the matrix useless for finding any sort of solution.

You can switch columns, but that doesn't get you any closer to a solution here.

If I were trying to find a solution, I might
  switch rows 1 and 2
  multiply the new row 1 by -3 and add it to the new row 2
  multiply the new row 1 by 2 and add it to row 3
This sequence of operations will make the first column [1 0 0], reducing the problem to 2×2 from 3×3.
3 0
3 years ago
Read 2 more answers
HELP PLEASE!!!!! WILL MARK AS BRAINLIEST!!!!
svlad2 [7]

The two sides marked 7 are congruent.

Sides AC and AD are congruent.

Sides BC and ED are congruent.

By SSS, the triangles are congruent.

Also, angles BAC and EAD are congruent, so by SAS, the triangles are congruent.

Answer: A. yes, by either SSS or SAS

5 0
3 years ago
Read 2 more answers
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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