All categories

3 years ago

A teacher's age is 6 years greater than 2 times a student's age. A principal's age is 10 years greater than 3 times

Mathematics

1 answer:

dlinn [17]3 years ago

5 0

Answer:

b. x + 4

Solution:

Student's age = x

Teacher's age = 6 + 2x

Principal's age = 10 + 3x

Difference in the age of the teacher and principal

(10 + 3x) - (6 + 2x)

= 10 + 3x - 6 - 2x

= (10 - 6) + (3x - 2x)

= 4 + x

You might be interested in

Find the number one-third of the way from 1/6 to 4/5.

masya89 [10]

I got u with this it really easy

3 0

3 years ago

Divide: 15m^2n + 20mn/5mn

mylen [45]

The answer is
15m^2n + 4/n
Hope that helps, have a nice night or day <4

8 0

3 years ago

Proving triangles similar

goldfiish [28.3K]

Answer:

x=11

Step-by-step explanation:

You do 38/3x+3 and 19/x+7 and then cross mulitply and get 57x+57=38x+266. Then yoy subtract 57 from 266 and get 57x=38x+209. Now you have to subtract 38 from 57 and then answer will be 19. So now you have 19x=209. Finally you divide 209 by 19 and get x=11. Good luck!

6 0

3 years ago

What is the solution to the system of equations? 7x + 3y = -11 2x + 5y=1

marta [7]

Answer:

x = -2 and y = 1

Step-by-step explanation:

for details see image

7 0

3 years ago

How to do this problem??

lorasvet [3.4K]

We know that the sum of the inner angles of any triangle is 180º

72º + (7x + 3)º + (3x + 5)º = 180º
72º + 7xº + 3º + 3xº + 5º = 180
7xº + 3xº = 180º - 72º - 3º - 5º
10xº = 100º
$x^0 = \frac{100^0}{10^0}$
$x^0 = 10\to\:\boxed{\boxed{x = 10^0}}\end{array}}\qquad\quad\checkmark$

The sum of the external angle (9y + 1)º with inner angle (3x + 5) = 180 °, Replace the measure of "x" found:

(9y + 1)º + (3x + 5)º = 180º
9yº + 1º + 3xº + 5º = 180º
9yº + 1º + 3.(10)º + 5º = 180º
9yº + 1º + 30º + 5º = 180º
9yº = 180º - 1º - 30º - 5º
9yº = 144º
$y^0 = \frac{144^0}{9^0}$
$y^0 = 16\to\:\boxed{\boxed{y = 16^0}}\end{array}}\qquad\quad\checkmark$

Answer:
The measures of "x" and "y" are respectively: 10º and 16º

6 0

3 years ago