Working with fractions, huh? They are not as hard as you may think.
Just think:
1/4 = 2/8
3/4= 6/8
So...
9 3/4 = 9 6/8
Step 2.
9 + 1 = 10
6/8 + 1/8 = 7/8
Step 3
10 and 7/8
Step 4
Simplest answer
10 7/8,
(can't get more simplified than that)
Hope this helps! :)
Answer:

We can rewrite this expression like this:

And we can use the quadratic formula given by:

Where 
And replacing we got:

And solving we got:

And since the value can't be negative the answer would be x = 32.75 and the value of y = 32.75+10 =42.75
Step-by-step explanation:
For this case we know that we have a rectangular playground and the area can be founded with this formula:

Where x represent the width and y the length. From the problem we know that A =1400 m^2 and the heigth is 10m longer than the wide so we can write this condition as:

And replacing this formula into the area we got:

We can rewrite this expression like this:

And we can use the quadratic formula given by:

Where 
And replacing we got:

And solving we got:

And since the value can't be negative the answer would be x = 32.75 and the value of y = 32.75+10 =42.75
Answer: (upside down fancy u) q5
Step-by-step explanation:
Simply apply the law of conservative (upside down fancy u)
Answer:
x = 5.5
Step-by-step explanation:
PM = MR
4x - 12 = -2x + 21 {add 2x to both the sides}
4x - 12 + 2x = -2x + 2x + 21
6x - 12 = 21 {Add 12 to both sides}
6x - 12 + 12 = 21 + 12
6x = 33
x = 33/6
x = 5.5
<h2>
The area of a triangle is =54 square units</h2><h2>
The perpendicular distance from B to AC is = 
</h2>
Step-by-step explanation:
Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

The area of a triangle is= ![\frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Bx_1%28y_2-y_3%29%20%2Bx_2%20%28y_3-%20y_1%29%2Bx_3%28y_1-y_2%29%5D)
=![|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|](https://tex.z-dn.net/?f=%7C%5Cfrac%7B1%7D%7B2%7D%20%5B2%282-8%2B12%288-1%29%2B12%281-2%29%5D%7C)
=
= 54 square units
The length of AC = 
= 
=
units
Let the perpendicular distance from B to AC be = x
According To Problem

⇔
units
Therefore the perpendicular distance from B to AC is = 