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2 years ago

5u-8v+9(2u-3v)-4(2v-5u)

Mathematics

2 answers:

Schach [20]2 years ago

7 0

Answer:

43u

Step-by-step explanation:

1

Distribute

5u-8v+9(2u-3v)-4(2v-5u)

5u-8v+18u-27v-4(2v-5u)

2

Rearrange terms

5u-8v+18u-27v-42v-5u)

5u-8v+18u-27v-4(-5u+2v)

3

Distribute

5u-8v+18u-27v-4(-5u+2v)

5u-8v+18u-27v+20u-8v

4

Combine like terms

5u-8v+18u-27v+20u-8v

43u-8v-27v-8v

5

Combine like terms

43u-8v-27v-8v

43u-43u

Elden [556K]2 years ago

5 0

Answer:

The answer is 43u – 33v

Step-by-step explanation:

5u – 8v + 9(2u – 3v) -4(2v – 5u)

5u – 8v + 18u – 27v – 8v + 20u

5u + 18u + 20u – 8v – 27v – 8v

43u – 33v

Thus, The answer is 43u – 33v

-TheUnknownScientist 72

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Step-by-step explanation:

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3 years ago

Read 2 more answers

A 2000 gallon fish aquarium can be filled by water flowing at a constant rate in 10 hours. When a decorativve rock is placed in

Elza [17]

Answer:

The volume of the rock is 13.368 cubic feet.

Step-by-step explanation:

The volume of the rock ( $V_{r}$ ), measured in gallons, is equal to the volume of the fish aquarium without the decorative rock ( $V'$ ) minus the volume of the fish aquarium with the decorative rock ( $V''$ ), both measured in gallons.

Since the water flow is at constant rate, the volume of the rock is expressed by the following equation:

$V_{r} = V' - V''$

$V_{r} = \dot V \cdot (\Delta t_{1}-\Delta t_{2})$ (1)

Where:

$\dot V$ - Flow rate, measured in gallons per hour.

$\Delta t_{1}$ - Filling time of the fish aquarium without the decorative rock, measured in hours.

$\Delta t_{2}$ - Filling time of the fish aquarium with the decorative rock, measured in hours.

And the flow rate is:

$\dot V = \frac{2000\,gal}{10\,h}$

$\dot V = 200\,\frac{gal}{h}$

The flow rate is 200 gallons per hour.

If we know that $\dot V = 200\,\frac{gal}{h}$ , $\Delta t_{1} = 10\,h$ and $\Delta t_{2} = 9.5\,h$ , then the volume of the rock is:

$V_{r} = \left(200\,\frac{gal}{h}\right)\cdot (10\,h-9.5\,h)$

$V_{r} = 100\,gal$

$V_{r} = 13.368\,ft^{3}$

The volume of the rock is 13.368 cubic feet.

8 0

3 years ago

The managers of 21 supermarkets counted the number of cars in their parking lots on the same day. The results are shown in the l

mixer [17]

The IQR is 42.5

Step-by-step explanation:

Interquartile range is the difference of third and first quartile.

First of all we have to find the median for that purpose the data has to be arranged in ascending order. The data is already in ascending order.

As the number of values are odd

n=21

The median will be: $(\frac{n+1}{2}) th\ term$

Putting n=21

$(\frac{21+1}{2})th\ term\\=(\frac{22}{2})th\ term\\= 11th\ term$

The 11th term is 133

So median = 133

Now the data is divided into two halves

One is: 98, 100, 101, 102, 108, 109, 111,118, 129, 132

2nd is: 135, 135, 145, 146, 146, 156, 170 176, 180, 180

Q1 will be the median of first half and Q3 will be the median of 2nd half.

As now the halves contain even number of values, the medians will be the average of middle two values

For First Half:

98, 100, 101, 102, 108, 109, 111,118, 129, 132

$Q_1 = \frac{108+109}{2}\\Q_1 = \frac{217}{2}\\Q_1 = 108.5$

For Second Half:

135, 135, 145, 146, 146, 156, 170 176, 180, 180

$Q_2 = \frac{146+156}{2}\\Q_2 = \frac{302}{2}\\Q_2 = 151$

Now

Interquartile Range:

$IQR = Q_3-Q_1\\= 151-108.5\\=42.5$

Hence,

The IQR is 42.5

Keywords: Median, IQR

Learn more about median at:

#LearnwithBrianly

7 0

3 years ago

What is 4 divided by 572

azamat

4/572

divide top and bottom by 4

1/143

or in decimal form

.006993007

4 0

3 years ago

How to do this question? : (

Lostsunrise [7]

How to do what question

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3 years ago

5u-8v+9(2u-3v)-4(2v-5u)​

5u-8v+9(2u-3v)-4(2v-5u)