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1 year ago

How many positive integers less than 1000 a) are divisible by 7? b) are divisible by 7 but not by 11?

Mathematics

1 answer:

sashaice [31]1 year ago

3 0

Answer:

130

Step-by-step explanation:

The number of integers meeting the criteria can be found by counting them using a counting formula.

<h3>Divisible by 7</h3>

Integers divisible by 7 will have the form (7n), where n is some positive integer. The number of them less than 1000 can be found from ...

7n ≤ 1000

n ≤ 142.857

There are 142 integers less than 1000 that are divisible by 7.

<h3>Divisible by 7 and 11</h3>

Similarly, integers divisible by 7 and 11 will be of the form (77n), for some positive integer n.

77n ≤ 1000

n ≤ 12.987

There are 12 integers less than 1000 that are divisible by both 11 and 7.

<h3>Divisible by 7, not 11</h3>

The number of integers less than 1000 that are divisible by 7, but not 11, will be the difference of these numbers.

142 -12 = 130 integers divisible by 7, but not 11.

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Solve the system of equations using matrices. Use the Gauss- Jordan elimination method And find a solution set

KATRIN_1 [288]

$\begin{gathered} x+y+z=4 \\ x-y-z=0 \\ x-y+z=8 \\ \text{The system using matrix is} \\ \begin{bmatrix}{1} & {1} & {1}, & {4} \\ {1} & {-1} & {-1,} & {0} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix}\rightarrow F2=F2-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix} \\ \rightarrow F3=F3-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F2=-\frac{1}{2}F2 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F3=F3+2F2=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {2,} & {8}{}{}\end{bmatrix}\rightarrow F3=\frac{1}{2}F3 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F2=F2-F3=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \rightarrow F1=F1-F3=\begin{bmatrix}{1} & {1} & {0}, & {0} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F1=F1-F2 \\ =\begin{bmatrix}{1} & {0} & {0}, & {2} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \text{Therefore, the solution is }x=2,\text{ y=-2 and z=4} \end{gathered}$

6 0

1 year ago

Factor completely x^3 + 1/8

maksim [4K]

Answer:

((2 x + 1) (4 x^2 - 2 x + 1))/8

Step-by-step explanation:

Factor the following:

x^3 + 1/8

Put each term in x^3 + 1/8 over the common denominator 8: x^3 + 1/8 = (8 x^3)/8 + 1/8:

(8 x^3)/8 + 1/8

(8 x^3)/8 + 1/8 = (8 x^3 + 1)/8:

(8 x^3 + 1)/8

8 x^3 + 1 = (2 x)^3 + 1^3:

((2 x)^3 + 1^3)/8

Factor the sum of two cubes. (2 x)^3 + 1^3 = (2 x + 1) ((2 x)^2 - 2 x + 1^2):

((2 x + 1) ((2 x)^2 - 2 x + 1^2))/8

1^2 = 1:

((2 x + 1) ((2 x)^2 - 2 x + 1))/8

Multiply each exponent in 2 x by 2:

((2 x + 1) (2^2 x^2 - 2 x + 1))/8

2^2 = 4:

Answer: ((2 x + 1) (4 x^2 - 2 x + 1))/8

5 0

3 years ago

Read 2 more answers

pioneers are buying mules to trek across the country. each of the 5 wagons require e mules to pull them. write the equation to e

tekilochka [14]

Answer:

d = 5e

Step-by-step explanation:

Number of wagons = 5

Number of miles per wagon = e

Total number of mules = d

Hence, relationship between total mules d needed will be ;

Total mules = number of wagons * Numbe rof mules per wagon

d = 5e

4 0

3 years ago

14) 25 , 13, 4, 20,16,3,2,5<br>

Igoryamba

Answer:

2,3,4,5,13,20,25 or

2,4,16,20(even number)

3,5,13,25 (odd number)

Step-by-step explanation:

6 0

3 years ago

Write an equation of the line that passes through (−1,3) and is parallel to the line y=2x+2

Komok [63]

Start off by writing the equation y = mx - b. Then plug in what you do know

3 = 2(-1) - b (m is the slope and parallel lines have the same slope). Then solve

3 = -2 - b

5 = b

So y = 2x + 5

6 0

3 years ago