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lawyer [7]
3 years ago
7

Please hurry its for a test simplify 1/3x + 1/4 + 2/3x

Mathematics
2 answers:
kykrilka [37]3 years ago
6 0

Answer:

x + 1/4

Step-by-step explanation:

Mama L [17]3 years ago
6 0

Answer:

X + 1/4

Step-by-step explanation:

id k I used mathpapa. it's a website for algebra.

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Emery earns money mowing his neighbors' lawns. He charges $1.25 per square yard for each lawn he mows. His neighbors, the Flores
aleksandr82 [10.1K]

Answer: 4.2 yards

Step-by-step explanation:

Let x = length of the Flores' lawn in yards.

Then, Area of lawn = 4x  sq. yard [Area of rectangle = width x length , here width = 4 yds (given)]

Charge per square yard = $1.25

Charge for Flores lawn = 1.25 × (Area of lawn)

= 1.25 (4x)

= 5x

As per given,

5x=21\\\\\Rightarrow\ x=\dfrac{21}{5}\\\\\Rightarrow\ x=4.2

Hence, the length of the Flores's lawn = 4.2 yards

5 0
3 years ago
C. A share of stock XYZ went from $40 a to $42 &amp; What was the dollar gain for the<br> stock XYZ?
Afina-wow [57]

Answer:

$2 was the dollar gain for the stock XYZ

5 0
3 years ago
????????????????? I NEED HELP WITH THIS WILL GIVE BRAINLIEST
GuDViN [60]

Answer:

#3 is 40

#4 is 90

Step-by-step explanation:

3 0
3 years ago
A shelf contains n separate compartments. There are r indistinguishable marbles.
densk [106]

Answer:\frac{n!}{r!(n-r)!}

ways

Step-by-step explanation:

Given that a shelf contains n separate compartments. There are r indistinguishable marbles

The marbles are identical so they can be placed in any order.

Let us consider the places available for placing these r marbles

No of compartments available =n

Marbles to be placed = r

Since marbles are identical and order does not matter

number of ways the r marbles can be placed in the n compartments

= nCr

=\frac{n!}{r!(n-r)!}

8 0
4 years ago
Consider a nine -year moving average used to smooth a time series that was first recorded in 1961. a. Which year serves as the f
kirill [66]

Answer:

  a.  1965

  b.  8 years

Step-by-step explanation:

For answers to questions like these it can work to consider the smallest possible dataset.

<h3>Dataset</h3>

For our purpose, consider the 9 years/values to be averaged to be ...

  1, 2, 3, 4, 5, 6, 7, 8 ,9

<h3>Observations</h3>

a. The center value of the data set is 5. Its number is 5-1=4 more than the first one.

The first centered value is from the year 1961 +4 = 1965.

b. The 4 values at the beginning, and the 4 values at the end do not have a corresponding "average" value. That is, 4+4 = 8 values in the series are lost with respect to the number of average values.

8 years of values are lost.

5 0
2 years ago
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