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ss7ja [257]
3 years ago
6

Which method do you prefer to use to find sums__count by tens and ones,use compatible numbers,or use friendly numbers and adjust

? Explain why.
Mathematics
1 answer:
DiKsa [7]3 years ago
5 0
I count by tens and ones. Let me show you why with an example;

X=56+27

What is x?

To solve this equation, all you have to do is add 56 and 27. Easy. Since there are 7 tens, (5 plus 2), 7 will be in the tenths place. Like this: 7_. What will be next to it? To find this, just add the ones. 6+7=13. “Uh. Oh, 13 is bigger than ten,) you might be wondering. This problem has a simple fix. How many tens are in 13? One. You add ten to your 7, and you now have 8 in your tens place. Add the remaining three ones, and you’ve got your answer. 83.

56+27=83

X=83.

This method is easier and faster than any other method. Did you know calculators are actually programmed to use this method for simple addition? Who would’ve thought?


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Find the midpoint of JK if J(-7. -5) and K(-3,7).
qaws [65]

Answer:

( − 5 , 1 )

Step-by-step explanation:

Use the midpoint formula to find the midpoint of the line segment.

7 0
3 years ago
Could I have some help on my patterns assignment​
nydimaria [60]

Answer:

1. 13, 18, 23, 28, 33, 38, 43, 48, 53

<em>Conjecture:</em> Each term is 5 more than the previous term.

2. 512, 256, 128, 64, 32, 16, 8, 4, 2

<em>Conjecture:</em> Each term is half of the previous term.

3. 1, 8, 27, 64, 125, 216, 343, 512, 729

<em>Conjecture:</em> Each term is a cube. To find the nth term, cube n.

4. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31

<em>Conjecture:</em> Each term is the next consecutive prime number.

5. <u><em>Correct as is</em></u>

6. 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89

<em>Conjecture:</em> Each term is the sum of the 2 previous numbers.

Step-by-step explanation:

1.

Given: 13, 18, 23, 28

<em>We add 5 to each number</em>

13 + 5 = 18

18 + 5 = 23

23 + 5 = 28

28 + 5 = 33

33 + 5 = 38

38 + 5 = 43

43 + 5 = 48

48 + 5 = 53

2.

Given : 512, 256, 128, 64

<em>Divide each number by 2</em>

512 ÷ 2 =256

256 ÷ 2 = 128

128 ÷ 2 = 64

64 ÷ 2 = 32

32 ÷ 2 = 16

16 ÷ 2 = 8

8 ÷ 2 = 4

4 ÷ 2 = 2

3.

Given: 1, 8, 27, 64

<em>We have to cube n:</em>

<u><em>n³ = consecutive number cubed</em></u>

<u><em></em></u>

1³ = 1 or 1 × 1 × 1

2³ = 8 or 2 × 2 × 2

3³ = 27 or 3 × 3 × 3

4³ = 64 or 4 × 4 × 4

5³ = 125 or 5 × 5 × 5

6³ = 216 or 6 × 6 × 6

7³ = 343 or 7 × 7 × 7

8³ = 512 or 8 × 8 × 8

9³ = 729 or 9 × 9 × 9

4.

Consecutive prime number: <u><em>Consecutive prime numbers are those that have no gaps or prime numbers between them.</em></u>

Look at the picture in the link:

6.

Given: 1, 1, 2, 3, 5, 8

<em>We must add the two previous numbers to get the next term:</em>

1 + 1 = 2

1 + 2 = 3

2 + 3 = 5

3 + 5 = 8

5 + 8 = 13

8 + 13 = 21

13 + 21 = 34

21 = 34 = 55

34 + 55 = 89

8 0
1 year ago
Pls Help! It would be very much appreciated, Thx!
Fofino [41]

Answer:

A. 30 two-point questions and 10 four-point questions

Step-by-step explanation:

30 multiplied by 2 equals 60

10 multiplied by 4 equals 40

60 plus 40 equals 100 points which is the amount of points the test is worth as stated i the question.

30 plus 10 equals 40, which is the total amount of questions in the test as stated in the question.

5 0
2 years ago
What are the roots of the following polynomial equation? x^3 + 72= 5x^2 + 18x (write from least to greatest)
kow [346]

Answer:

2:3

Step-by-step explanation:

8 0
2 years ago
Find the simple interest and amount when principal = Rs. 6000, rate = 6% per annum and time = 4 years
Alinara [238K]

Given:

Principal = Rs. 6000

Rate of simple interest = 6% per annum.

Time = 4 years

To find:

The simple interest and amount.

Solution:

Formula for simple interest:

I=\dfrac{P\times r\times t}{100}

Where, P is principal, r is the rate of interest and t is the number of years.

Putting P=6000, r=6 and t=4, we get

I=\dfrac{6000\times 6\times 4}{100}

I=60\times 6\times 4

I=1440

Now,

Amount=Principal+Interest

Amount=6000+1440

Amount=7440

Therefore, the simple interest is Rs. 7440 and the amount is Rs 7440.

7 0
2 years ago
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