If 26.97 ÷ 6.3 is written in long division form as long division set up where 63 is on the outside of the division symbol and 26970 is on the inside so that the divisor is written as a whole number, where should the decimal point be placed in the dividend?
Between the 2 and the 6
Between the 6 and the 9
Between the 9 and the 7
Between the 7 and the 0

Answers

Answer 1

The decimal point should be placed between digits 9 and 7 in the dividend. i.e the dividend becomes 269.7

Step-by-step explanation:

Dividend = 26.97 [numerator]

Divisor = 6.3 [denominator]

If 26.97 ÷ 6.3 is written in long division form so that the divisor is written as a whole number, we have the following;

(i)First convert the divisor to a whole number by multiplying by 10 i.e

6.3 x 10 = 63

(ii) Since the divisor (denominator) has been multiplied by 10, to make sure the division expression stays the same, we need to multiply the dividend(numerator) too by 10. i.e

26.97 x 10 = 269.7

(iii) The division expression then becomes;

269.7 ÷ 63

Therefore, the decimal point should be placed between digits 9 and 7 in the dividend.

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Answer 2

Answer:

9 and 7

Step-by-step explanation:

26.6.3 is long division set up 63 is out of the division symbol and 26970.

Related Questions

y=x-12 y=13 solve for x

Answers

Answer:

Step-by-step explanation:

x - 12 = 13

x = 25

y = 12

Answer:

x=25

Step-by-step explanation:

y=x-12....eq1

y=13.... eq2

Substituting eq2 in 1:-

13=x-12

x=12+13=25

a. Write 60,000,000 in words.

Answers

Answer:

The answer is sixty million.

How many different letter permutations, of any length, can be made using the letters M O T T O (For instance, there are 3 possible permutations of length 1.)

Answers

Answer:

Step-by-step explanation:

Given that

2 O, 2 T and 1 M

Now based on this, the following ways are there

Three ways i.e. {M,T,O}

XX or XY

XX in 2C1 = two ways i.e. {OO or TT}

XY in 3C2 × 2! = six ways

XXY or XYZ

XXY in 2C1 × 2C1 × 3! ÷ 2! = twelve ways

XYZ in 3C3 × 3! = six ways

XXYY or XXYZ

XXYY = 4! ÷ (2! × 2!) = six ways

XXYZ in 2C1 × 4! ÷ 2! = twenty four ways

= 5! ÷ (2! ×2!)

= 120 ÷ 4

= 30

Therefore, the total is

= 3 + (2 +6)+ (12 +6) + (6 +24) + 30

= 89

There are 30 different letter permutations.

The word is given as:
MOTTO

From the given word, we have the following parameters:

[tex]n = 5[/tex] --- the number of characters in the word[tex]T = 2[/tex] --- the number of times letter T appears[tex]O = 2[/tex] --- the number of times letter O appears

The number of different (d) letter permutation is then calculated as:

[tex]d = \frac{n!}{T!O!}[/tex]

This gives

[tex]d = \frac{5!}{2!2!}[/tex]

Evaluate each factorial

[tex]d = \frac{120}{2 \times 2}[/tex]

Evaluate the product of 2 and 2

[tex]d = \frac{120}{4}[/tex]

Divide 120 by 4

[tex]d = 30[/tex]

Hence, there are 30 different letter permutations.

Answers

Answer:

Step-by-step explanation:

A complex number is a number that has both real and imaginary part. An example of a complex number is z = x+iy where x is the real part and y is the imaginary part.

If z = x+iy, the conjugate of z₁ will be x-iy

adding the complex number to its conjugate will give;

P = z+z₁

P = x+iy +(x-iy)

open the parenthesis

P = x+iy+x-iy

collect like terms

P = x+x+iy-iy

P = 2x

We can see that the resulting value does not contain the imaginary number i, hence the result is a real number.

Hence the result of adding a complex number to its conjugate is a REAL NUMBER

Taking the difference;

P = = z-z₁

P = x+iy -(x-iy)

open the parenthesis

P = x+iy-x+iy

collect like terms

P = x-x+iy+iy

P = 0+2iy

P = 2iy

We can see that the resulting value contains the imaginary number i, hence the of taking their difference is a complex number.

Hence the result of subtracting a complex number from its conjugate is an IMAGINARY NUMBER

For the function f(x) = 16x^3 - x, make a table of slopes of secant lines and make a conjecture about the slope of the tangent line at x = 1. Complete the table. (Round the final answer to three decimal places as needed. Round all intermediate values to four decimal places as needed.) An accurate conjecture for the slope of the tangent line at x = 1 is (Round to the nearest integer as needed.)

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

Interval Slope of secant lines

[1,2] 113.000

[1, 1.5] 75.000

[1, 1.1 ] 51.960

[1, 1.01] 47.482

[1, 1.001] 47.048

Step-by-step explanation:

From the question we are told that

The function is [tex]f(x) = 16x^3 -x[/tex]

Now considering the intervals

For [1, 2]