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3 years ago

Please answer quickly!!!

Mathematics

1 answer:

Mademuasel [1]3 years ago

4 0

Answer:

1. 64cm² , 2. 240yd² , 3.≈220.5cm² , 4.≈193.4m² , 5. 38.6ft² , 6. ≈78.1yd² , 7. ≈120.7

Step-by-step explanation:

1. A=12x4.5+2x5 , 2. A=24x8+0.5x12x8 3. A=0.5x16x15+0.5x8²π ,4. A=7x15+0.5(15/2)² 12 ,5. A=3.6(16/2)+0.5x7x2.8 ,6. A=8(16/2)+0.5x9π ,7. A=(8/4)x5²xcot(180/8)

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Count by fives from 135!to 175 describe the pattern

irina1246 [14]

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2 years ago

Ne #3: When I take half of my number and add two, I get twenty-four. What is my number

irga5000 [103]

Answer:

My number will be 44.

Step-by-step explanation:

My number will be divided by 2.
I will get 22.
Then I'll 2 to my 22.
It will give me 24.

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4 0

3 years ago

Miki has 104 nickels and 88 dimes. She wants to divide her coins into groups where each group has the same number of nickels and

pochemuha

Given :

Miki has 104 nickels and 88 dimes.

She wants to divide her coins into groups where each group has the same number of nickels and the same number of dimes.

To Find :

Largest number of groups she can have .

Solution :

In the given question we need to find the largest number of groups she can have i.e we have to find the LCM of 104 and 88 .

Now , factorizing both of them , we get :

$104=1\times 2\times 2\times 2\times 13$

$88=1\times 2\times 2\times 2\times 11$

Form above , we can say that common factors are :

$HCF=2\times 2\times 2=8$

Therefore , the largest number of groups she can have is 8 .

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6 0

3 years ago

Read 2 more answers

205 students enrolled in a freshman-level chemistry class. By the end of the semester, 4 times the number of students passed as

AlexFokin [52]

Answer: 164 students passed and 41 students failed.

Step-by-step explanation:

Let the number of students be x

x students failed, but 4 times the failed students passed.

So 4x students passed and x students failed.

4x + x = 205

5x = 205

x = 205/5

x = 41

41 times 4 is 164 and 41 times 1 is 41.

So 164 students passed and 41 students failed

3 0

2 years ago

What is the probability that no 1s will be in the output for an input sequence of 10 digits??

emmasim [6.3K]

The total numbers with 10 digits are 9*10*10*10*10*10*10*10*10*10 = 9*(10^9)

If you are restricted to use no 1s, the numbers are 8*(9^9)

Then, the probability is : numbers with no 1s / total numbers

= 8*(9^9) / [9*(10^9)]= 0.344

5 0

3 years ago