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sashaice [31]
3 years ago
11

On a scale drawing, the height of a tree is 2.25 inches.

Mathematics
2 answers:
Ostrovityanka [42]3 years ago
8 0

Answer:

It's D. 2.25*75=168.75

Alexeev081 [22]3 years ago
7 0

Answer:

168.75

Step-by-step explanation:

Since the height of the tree is 2.25.

2.25×75= 168.75

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Please help thank you
aivan3 [116]

Answer:

a = 4

b = 11

Step-by-step explanation:

To convert the recurring decimals into fraction we will follow the following process.

Given recurring decimal is 0.36363636....

Let x = 0.363636......... (1)

Multiply this decimal with 100.

100x = 36.36363636...........(2)

Now subtract expression (1) from expression (2)

100x = 36.363636.......

<u>      x =   0.363636.......</u>

99x = 36

x = \frac{36}{99}

x = \frac{4}{11}

Since, x=\frac{a}{b}=\frac{4}{11}

a = 4 and b = 11 will be the answer.

4 0
3 years ago
5 ft
Rom4ik [11]

Answer:

Assuming this is a rectangular prism the surface area would be 324ft^2.

Step-by-step explanation:

Surface Area = 2(lw + lh + wh).

3 0
3 years ago
Read 2 more answers
0.579 x 10 by the power of 4 <br>correct answer gets brainliest
MrRa [10]
Regular notation is 5790. scientific notation is 5.79 x 10^3
5 0
3 years ago
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You and three friends start a small business the total income is 1,820 and the total expenses are 360 you share the profit evenl
Verizon [17]

Answer:

Each of them get 365.

Explanation:

Total income =  1820

Total expense = 360

Total profit = Total income - Total expense = 1820 - 360 = 1460

Number of persons sharing the profit = 4

Profit got by each person = Total profit/ number of persons sharing profit

                           = 1460/4 = 365

So each of them get 365.

6 0
4 years ago
In a box of 10 calculators, one is defective. In how many ways can four calculators be selected, if you know one
Juliette [100K]

Given:

Total number of calculators in a box = 10

Defective calculators in the box = 1

To find:

The number of ways in which four calculators be selected and one  of the four calculator is defective.

Solution:

We have,

Total calculators = 10

Defective calculators = 1

Then, Non-defective calculator = 10-1 = 9

Out of 4 selected calculators 1 should be defective. So, 3 calculators are selected from 9 non-defective calculators and 1 is selected from the defective calculator.

\text{Total ways}=^9C_3\times ^1C_1

\text{Total ways}=\dfrac{9!}{3!(9-3)!}\times 1

\text{Total ways}=\dfrac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}

\text{Total ways}=\dfrac{9\times 8\times 7}{6}

\text{Total ways}=3\times 4\times 7

\text{Total ways}=84

Therefore, the four calculators can be selected in 84 ways.

7 0
3 years ago
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