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lakkis [162]
3 years ago
15

!HELP!

Mathematics
2 answers:
creativ13 [48]3 years ago
6 0
Let x=w, and then L=2x+3

Since area is LW we have:

A=x(2x+3)

A=2x^2+3x   (c.)
choli [55]3 years ago
4 0
X^2+3x

that be my answer
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A small business spend $23,000 for taxable items last year. The tax rate was 9.5%. How much did the small business pay in taxes
ioda

Answer:

$2,185

Step-by-step explanation:

Multiply the amount spent by the tax rate.

23,000 x 9.5% = $2,185

8 0
3 years ago
What’s the system of y=x-4 and y=4x-10?
Natasha_Volkova [10]

Answer:

That is the system.

y = x - 4

y = 4x - 10

If you are looking for the solution then use the method of substitution or the method of elimination.

4 0
3 years ago
Read 2 more answers
The mean of seven positive integers is 16. When the smallest of these seven integers is removed, the sum of the remaining six in
VashaNatasha [74]

Answer: The value of the integer that was removed = 4

Step-by-step explanation:

Sum of <em>n</em> values= Mean x <em>n</em>

Given: n = 7 , Mean = 16

Let x= integer removed.

Sum of remaining 6 integers = 6

\Rightarrow\ x+108 = 16\times7\\\\\Rihghtarrow\ x+108= 112\\\\\Rightarrow\ x=112-108\\\\\Rightarrow\ x= 4

Hence, the value of the integer that was removed = 4

8 0
3 years ago
True or false and any nonzero integer divided by zero equals zero
kiruha [24]
False, it would equal the nonzero integer
any nonzero integer MULTIPLIED by zero would equal zero
3 0
3 years ago
The sum of four consecutive odd integers is three more than five times the least of the integers. Find the integers.
Olegator [25]

Answer:

<h3>              9, 11, 13, 15</h3>

Step-by-step explanation:

{k - some integer}

2k+1  - the first odd integer (the least)

5(2k+1)  - five times the least

5(2k+1)+3 -<u> three more than five times the least</u>

2k+1+2 = 2k+3  - the odd integer consecutive to 2k+1

2k+3+2 = 2k+5  - the next odd consecutive integer (third)

2k+5+2 = 2k+7  - the last odd consecutive integer (fourth)

2k+1+2k+3+2k+5+2k+7 - <u>the sum of four odd consecutive integers</u>

2k+1 + 2k+3 + 2k+5 + 2k+7 = 5(2k+1) + 3

8k + 16 = 10k + 5 + 3

     - 10k       -10k

-2k + 16 = 8

     -16       - 16    

      -2k = -8  

    ÷(-2)    ÷(-2)  

      k = 4

2k+1 = 2•4+1 = 9

2k+3 = 2•4+3 = 11

2k+5 = 2•4+5 = 13

2k+7 = 2•4+7 = 15  

Check: 9+11+13+15 = 48;  48-3 = 45;  45:5 = 9 = 2k+1

4 0
3 years ago
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