Answer:
The numbers are 12 and 3.
Step-by-step explanation:
We can solve this problem by working with the information we have and setting up some equations.
We know that one number is four times as large as another. So, let the smaller number be represented by the variable x and the bigger number be represented by 4x, since it is four times as large.
Now, we know that if the numbers are added together, then the result is six less than seven times the smaller number. This can also be represented by the equation 4x + x = 7x - 6.
Let's solve that equation like so:

So, the smaller number must be 3 (remember that x represented the smaller number). To find the bigger number, all we need to do is multiply 3 by 4, which gives us 12. Therefore, the numbers are 12 and 3.
Ok
first of all, for q(x)/p(x)
if the degree of q(x) is less than the degree of p(x),then the horizontal assemtote is 0
then simplify
any factors you factored out is now a hole, remember them
to find the vertical assemtotes of a function, set the SIMPLIFIED denomenator equal to 0 and solve
so
y=(x-5)/(x^2-1)
q(x)<p(x)
horizontal assemtote is y=0
no factors to simplify so no holes
set denomenator to 0 to find vertical assemtote
x^2-1=0
(x-1)(x+1)=0
x-1=0
x=1
x+1=0
x=-1
the horizontal assemtotes are x=1 and -1
Answer:

Step-by-step explanation:
Given:

1.

2.

3.

4.

5.

6.

7.

8.

Answer:
y = -2x + 5
Step-by-step explanation:
To find the slope, you need to pick two points and put into the slope formula. It doesn't matter which ones, you will get the right answer regardless. I used (-3, 11) and (2, 1).

Now that you have the slope, you can plug it into point-slope form to find the equation in slope-intercept form. You will also need to plug one of the given points. Again, it doesn't matter which one.

Answer:
Area of play ground = Length × Breadth = 750 × 250 = 187500. a Cost of levelling the ground = 187500 × 16 100 = Rs . 30000