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1 year ago

9

Polygon v has an area of 1 square unit. Jun-seo drew a scaled version of polygon v using a scale factor of 7 and labeled it poly

gon w

2 answers:

Liula [17]1 year ago

8 0

Answer:

49 units

Step-by-step explanation:

khan academy

son4ous [18]1 year ago

3 0

Answer:

The area of Polygon

W

WW is

49

4949 square units.

Step-by-step explanation: dk

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A restaurant served 25,000 customers in 2012. In 2014, the restaurant served 26,600 customers. Write a linear model that represe

DerKrebs [107]

Answer:

Step-by-step explanation:

y=25000(1+6,4%)^x

x=8 (2012-2020)

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2 years ago

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Kari bought 3 boxes of cookies to share with a book club. Each box contains 12 cookies. The expression represents the number of

ivanzaharov [21]

<u>ANSWER:</u>

Kari bought 3 boxes of cookies to share. The algebraic expression is $\frac{36}{x}$

<u>Solution:</u>

Given, Kari bought 3 boxes of cookies to share with a book club.

Each box contains 12 cookies.

So, in total we have 3 x 12 cookies = 36 cookies.

Now, we have to find how many cookies can each person p will get.

Let, the total number of persons be x.

Then, after equally sharing the cookies,

$\text { Number of cookies with each person }=\frac{\text {number } of \text { cookies available.}}{\text {total number of persons}}$

$=\frac{36}{x} \text { cookies. }$

Hence, the algebraic expression is $\frac{36}{x}$

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2 years ago

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Roger went shoe shopping with his cousins Derek and Kamal. Roger bought a pair of basketball shoes that cost $74.95. Derek spent

bezimeni [28]

Answer:

I'm pretty sure its $19.65

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2 years ago

There are 11 portable mini suites (a.k.a. cages) in a row at the Paws and Claws Holiday Pet Resort. They are neatly labeled with

BigorU [14]

Step-by-step explanation:

a) 7!

If there are no restrictions, answer is 7! as it is the permutation of all animals.

b) 4! x 3!

As cats are 6 and Dogs are 5, thus 1st and last must be cats in order to have alternate arrangements. Therefore the only choices are the order of the cats among themselves and the order of the dogs among themselves. There are 4! permutations of the cats and 3! permutations of the dogs, so there are a total of 4! x 3! possible arrangements of the suites.

c) 3! x 5!

There are 3! possible arrangements of the dogs among themselves. Now, if we consider the dogs as one ”object” together, then we can think of arranging the 4 cats together with this 1 additional object. There are 5! such arrangements possible, so there are a total of 3! · 5! possible arrangements of the suites.

d) 2 x 4! x 3!

As required that all the cats must be together and all the dogs must be together, either the cats are all before the dogs or the dogs are all before the cats. There are two possible arrangements thus two times of both possibilities is the answer i.e. 2 x 4! x 3!

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2 years ago

The heights of the trees for sale at two nurseries are shown below. Heights of trees at Yard Works in feet : 7, 9, 7, 12, 5 Heig

Troyanec [42]

Answer:

The mean of tree heights at Yard Works is 8 feet and at Grow Station is 9 feet.The mean absolute deviation of the tree heights at both yards is 2.

Step-by-step explanation:

1). Height of the trees at Yard Works are = 7,9,7,12,5 feet

So mean height of the trees = (7+9+7+12+5)÷5

= 40÷5 =8 feet

Standard deviation of the trees at Yard works = ∑(║(height of the tree-mean height of the tree))║/(number of trees)

(height of the tree-mean height of the tree)= ║(7-8)║+║(9-8)║+║(7-8)║+║(12-8)║+║(5-8)║ = (1)+1+(1)+4+(3)= 10

Therefore standard deviation = (10)/(5) =2

2). In the same way mean height of the trees at Grow Station=(9+11+6+12+7)/5= 45/5 = 9

Now we will calculate the mean deviation of the tress at Grow Station

= ∑║(height of the tree-mean height of the tree)║/(number of trees)

= ║(9-9)║+║(11-9)║+║(6-9)║+║(12-9)║+║(7-9)║/(5)

= (0+2+3+3+2)/5

= 10/5 =2

Therefore The mean of tree heights at Yard Works is 8 feet and at Grow Station is 9 feet.The mean absolute deviation of the tree heights at both yards is 2.

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2 years ago

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