5x²+13x-4-17x²-7x+19+15x²+5x-21x-7=
=5x²+15x²-17x² +(13x-7x+5x-21x) -4-7+19=
=3x²-10x+8
1 Chicken have = 2 legs
1 Cow have = 4 legs
No of cows = x
No of chickens = y
Total no of legs = 4x + 2y = 210.............(1)
Total No of heads = x + y = 84................(2)
x = 84 - y....( from (2) )
Substituting the above equation in (1)
4 ( 84 - y) + 2y = 210
336 - 4y + 2y = 210
336 - 2y = 210
-2y = -126
= y = 63
No of cows = 84-63 = 21
∴ <u><em>Total no of cows = 21</em></u>
<u><em>Total no of chickens = 63</em></u>
It is the formula to find the longest line on a right triangle. for example, a squared plus b squared equals c squared. In this case, c must always be the line not relating to the right angel.
Answer:
Total number of tables of first type = 23.
Total number of tables of second type = 7
Step-by-step explanation:
It is given that there are 30 tables in total and there are two types of tables.
Let's call the two seat tables, the first type as x and the second type as y.
∴ x + y = 30 ......(1)
Also a total number of 81 people are seated. Therefore, 2x number of people would be seated on the the first type and 5y on the second type. Hence the equation becomes:
2x + 5y = 81 .....(2)
To solve (1) & (2) Multiply (1) by 2 and subtract, we get:
y = 7
Substituting y = 7 in (1), we get x = 23.
∴ The number of tables of first kind = 23
The number of tables of second kind = 7
Answer:
$848
Step-by-step explanation:
Calculation for the cost of this familiy's clothing in 1991
First step is to calculate the amount that is increase per year
Increase per year= ($1000-$620)/(1995-1985)
Increase per year= 380/10
Increase per year= $38
Now let y be :38*x + $620 and let x be 6 years (1991-1985)
Second step is to calculate the cost of the clothing in 1991
y = 38*6 + 620
y=228+620
y = $848
Therefore the cost of this familiy's clothing in 1991 will be $848
