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Aneli [31]
3 years ago
7

18 ÷ (-30) =???????????????

Mathematics
2 answers:
Basile [38]3 years ago
7 0
The answerrrr is -0.6
AlekseyPX3 years ago
6 0
The answer is -0.6.
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Find the square roots by division method of 210,681 please tell me
padilas [110]

Answer:

  459

Step-by-step explanation:

The "long division method" algorithm for square root makes use of the relation described by the square of a binomial.

  (a +b)² = a² +2ab +b² = a² +b(2a +b)

<h3>Steps</h3>

The value for which the root is desired is written with digits marked off in pairs either side of the decimal point.

The initial digit of the root is the integer part of the square root of the most-significant pair. Here that is floor(√21) = 4. This is shown in the "quotient" spot above the leftmost pair. The square of this value is subtracted, and the next pair brought down for consideration. Here, that means the next "dividend" is 506.

The next "divisor" will be 2 times the "quotient" so far, with space left for a least-significant digit. Here, that means 506 will be divided by 80 + some digits. As in regular long division, determining the missing digit involves a certain amount of "guess and check." We find that the greatest value 'b' that will give b(80+b) ≤ 506 is b=5. This is the next "quotient" digit and is placed above the "dividend" pair 06. The product 5(85) = 425 is subtracted from 506, and the next "dividend" pair is appended to the result. This makes the next "dividend" equal to 8181.

As in the previous step, the next "divisor is 2 times the quotient so far: 2×45 = 90, with space left for the least significant digit. 8181 will be divided by 900-something with a "quotient" of 9. So, we subtract the product 9(909) = 8181 from the "dividend" 8181 to get the next "dividend." That result is zero, so we're finished.

The root found here is 459.

__

<em>Additional comment</em>

In practice, roots are often computed using iterative methods, with some function providing a "starter value" for the iteration. Some iterative methods can nearly double the number of good significant digits in the root at each iteration.

Using this "long division method," each "iteration" adds a single significant digit to the root. Its advantage is that it always works, and is generally suitable for finding roots by hand. Once the number of root digits begins to get large, the "divisor" starts to be unwieldy.

8 0
2 years ago
Find the measures of the following angles
mezya [45]

Answer:

HGD: 117

FDG: 180

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
HELP ME PLEASE WITH 1,2 AND 3 PLEASE DON'T TAKE LONG TO RESPOND !!​
soldi70 [24.7K]

Answer:

1. -19x

3. 23 -  8x

5. -18x + 2y

Step-by-step explanation:

first: 3x - x = 2x

2x - 22x = -20x

-20x + x = -19x

second one:

8 - (-15) = 23

12x - 20x = -8 x

23 - 8x

third:

-10x - 8x = -18x

3y - y = 2y

-18x + 2y

6 0
3 years ago
Read 2 more answers
What kind of sequence is the pattern 1, 6, 7, 13, 20, ...?
AlexFokin [52]

Answer:

The given sequence 6, 7, 13, 20, ... is a recursive sequence

Step-by-step explanation:

As the given sequence is

6, 7, 13, 20, ...

  • It cannot be an arithmetic sequence as the common difference between two consecutive terms in not constant.

As

d = 7-6=1,  d = 13-7=6

As d is not same. Hence, it cannot be an arithmetic sequence.

  • It also cannot be a geometrical sequence and exponential sequence.

It cannot be geometric sequence as the common ratio between two consecutive terms in not constant.

As

r = 7-6=1,  r = 13-7=6

r = \frac{7}{6},  r = \frac{13}{7}

As r is not same, Hence, it cannot be a geometric sequence or exponential sequence. As exponential sequence and geometric sequence are basically the same thing.

So, if we carefully observe, we can determine that:

  • The given sequence 6, 7, 13, 20, ... is a recursive sequence.

Please have a close look that each term is being created by adding the preceding two terms.

For example, the sequence is generated by starting from 1.

     {\displaystyle F_{1}=1

and

     {\displaystyle F_{n}=F_{n-1}+F_{n-2}}

for n > 1.

<em>Keywords: sequence, arithmetic sequence, geometric sequence, exponential sequence</em>

<em>Learn more about sequence from brainly.com/question/10986621</em>

<em>#learnwithBrainly</em>

6 0
3 years ago
What is the range of the equation
a_sh-v [17]

The range of the equation is y>2

Explanation:

The given equation is y=2(4)^{x+3}+2

We need to determine the range of the equation.

<u>Range:</u>

The range of the function is the set of all dependent y - values for which the function is well defined.

Let us simplify the equation.

Thus, we have;

y=2 \cdot 4^{x+3}+2

This can be written as y=2^{1+2(x+3)}+2

Now, we shall determine the range.

Let us interchange the variables x and y.

Thus, we have;

x=2^{1+2(y+3)}+2

Solving for y, we get;

x-2=2^{1+2(y+3)}

Applying the log rule, if f(x) = g(x) then \ln (f(x))=\ln (g(x)), then, we get;

\ln \left(2^{1+2(y+3)}\right)=\ln (x-2)

Simplifying, we get;

(1+2(y+3)) \ln (2)=\ln (x-2)

Dividing both sides by \ln (2), we have;

2 y+7=\frac{\ln (x-2)}{\ln (2)}

Subtracting 7 from both sides of the equation, we have;

2 y=\frac{\ln (x-2)}{\ln (2)}-7

Dividing both sides by 2, we get;

y=\frac{\ln (x-2)-7 \ln (2)}{2 \ln (2)}

Let us find the positive values for logs.

Thus, we have,;

x-2>0

     x>2

The function domain is x>2

By combining the intervals, the range becomes y>2

Hence, the range of the equation is y>2

7 0
4 years ago
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