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3 years ago

Find the solution of the inequality -3n < 81

Mathematics

1 answer:

posledela3 years ago

8 0

Answer:

n<-27

Step-by-step explanation:

you have to get n by itself, to do this divide -3 from both sides of the equations.

-3n/-3<81/-3

n<-27

the answer is n<-27

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Find out net profit after tax from the following information:

WARRIOR [948]

Answer:

Net Profit after tax Rs 15,000

Step-by-step explanation:

The computation of the net profit after tax is shown below:

Gross profit Rs. 1,25,000

Less:

Selling and distribution expenses Rs. 21,000

General and administrative expenses Rs. 75,000

Interest on loan Rs. 5,000

Gain on sale of plant Rs. 4,000

Profit before tax Rs 20,000

Less: income tax expense at 25% of Rs 20,000 Rs 5000

Net Profit after tax Rs 15,000

3 0

3 years ago

one sixth of the students in a school marching band are in the percussion section, the percussion section has 6 students, how ma

Igoryamba

Answer:

36 students

Step-by-step explanation:

because if 1/6 is 6 students then you have to multiply the 6 by 6

4 0

3 years ago

Read 2 more answers

A triangle has an area of 54 square inches the height of the triangle is 12 inches what is the length of the base of the triangl

prohojiy [21]

Answer:

<em>The base of the triangle = 9inches</em>

Step-by-step explanation:

<u>Explanation</u>:-

The area of the triangle = $\frac{1}{2} bh$ square units

Given area of the triangle(A) = 54 square inches

Given the height of the triangle (h) = 12 inches

now equating $\frac{1}{2} b h = 54$

$\frac{1}{2} b (12) = 54$

now simplification, we get

6 b = 54

Dividing '6' on both sides, we get

b = 9

The base of the triangle = 9 inches

<u>Conclusion</u>:-

<em>The base of the triangle = 9 inches</em>

6 0

3 years ago

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
$7^{101}=7\times(7^{50})^2$
$7^{50}=(7^{25})^2$
$7^{25}=7\times(7^{12})^2$
$7^{12}=(7^6)^2$
$7^6=(7^3)^2$
$7^3=7\times7^2$

and so, using the property that, if $a_1\equiv b_1\mod n$ and $a_2\equiv b_2\mod n$ , then $a_1a_2\equiv b_1b_2\mod n$ , we get

$7\equiv3\mod4$
$7^2\equiv9\equiv1\mod4$
$7^3\equiv7\times1\equiv7\equiv3\mod4$
$7^6\equiv9\equiv1\mod4$
$7^{12}\equiv1\mod4$
$7^{25}\equiv7\times1\equiv7\equiv3\mod4$
$7^{50}\equiv9\equiv1\mod4$
$7^{101}\equiv7\times1\equiv7\equiv3\mod4$
$7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4$