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mr_godi [17]
3 years ago
11

You can get 4 value meals for 21.88 at the restaurant how much is each value meal?

Mathematics
2 answers:
GuDViN [60]3 years ago
3 0

Answer:

$5.47 each meal

Step-by-step explanation:

21.88 divided by 4 equals 5.47

antoniya [11.8K]3 years ago
3 0
The correct answer is $5.47
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What is the slope of f(x) = 2x – 3
bezimeni [28]

Answer:

2

Step-by-step explanation:

slope is 2 because is the coefficient is next to the variable x

6 0
3 years ago
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If you paint 1/6 of a fence each day, how many day would it take you to paint 3/4 of the fence?
ololo11 [35]
(3/4) / (1/6)

(3/4) * (6/1) = 18/4 = 4 2/4 = 4 1/2 days
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3 years ago
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Help with math Plsss I don't understand how to do this (I got this wrong but I have another try)​
Anuta_ua [19.1K]
<h3>Answer: 920</h3>

=================================================

Explanation:

The front face is a triangle with area of base*height/2 = 15*8/2 = 60 square feet.

The back face is identical to the front face, so we have another 60 square feet.

The left rectangular wall is 20 ft by 8 ft tall. Its area is 20*8 = 160 ft^2

The right slanted rectangular face is 20 ft by 17 ft. Its area is 20*17 = 340 ft^2

Lastly, the bottom rectangle floor is 15 ft by 20 ft to give an area of 15*20 = 300 ft^2

------------------

To summarize so far

  • Front face = 60 ft^2
  • Back face = 60 ft^2
  • Left face = 160 ft^2
  • Right slanted face = 340 ft^2
  • Bottom floor = 300 ft^2

Add up those areas to get the overall surface area.

60+60+160+340+300 = 920

6 0
2 years ago
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Simplify this expression 8+19m-6m
gladu [14]
I am pretty sure that it would be 13m + 8. All you have to do is combine the like terms.
7 0
3 years ago
What is the coefficient of the x5y5-term in the binomial expansion of (2x – 3y)10? 10C5(2)5(3)5 10C5(2)5(–3)5 –10C5(2)5(–3)5 10C
butalik [34]

ANSWER

10C_5(2)^{5}( - 3)^5

EXPLANATION

The given binomial expansion is:

{(2x - 3y)}^{10}

Compare this to

{(a + b)}^{n}

we have a=2x , b=-3y and n=10

We want to find the coefficient of the term

{x}^{5}  {y}^{5}

This implies that, r=5.

The terms in the expansion can be obtained using

T_{r+1}=nC_ra^{n-r}b^r

We substitute the given values to obtain;

T_{5+1}=10C_5(2x)^{10-5}(  - 3y)^5

T_{6}=10C_5(2x)^{5}(3y)^5

T_{6}=10C_5(2)^{5}( - 3)^5 {x}^{5}  {y}^{5}

Hence the coefficient is;

10C_5(2)^{5}( - 3)^5

5 0
3 years ago
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