Answer:
p = 2 ; q = 4
Step-by-step explanation:
Given tbe equation :
3p + 2q = 14 - - - (1)
10p + 6q = 44 - - -(2)
What is p and what is q
This is a simultaneous equation ; using elimination method :
Multiply (1) by 6 and (2) by 2
18p + 12q = 84 - - - - (3)
20p + 12q = 88 - - - (4)
Subtract (3) and (4)
-2p = - 4
p = 4/2
p = 2
Put p = 2 in (1)
3p + 2q = 14
3(2) + 2q = 14
6 + 2q = 14
2q = 14 - 6
2q = 8
q = 8/2
q = 4
p = 2 ; q = 4
Answer:
Firstly find the area of the square and then the area of the two triangles and then plus it. That's what I think I tried
Answer:
11 (if you wanna know how many cups [which i mean by 2/3] it is 1)
Step-by-step explanation:
Well, to start, a cook has 11 7/8 cups of flour, right? Well, if you subtract 11 7/8 by 2/3 you get 11 5/24. As an estimate, you would get just plain 11 because there is three numbers when you estimate, 11, 11 1/2, or 12. 11 5/24 is more closer to 11 (because 1/2 would be 11 12/24, and 12 would be 11 13/24 or more until 12)
Hope it helped...
Answer:


Step-by-step explanation:
Given

Solving (a): 
We have:

Express f(x) as y

Swap x and y

Add 8 to 


Square both sides

Rewrite as:

Express y as: 

To determine the domain, we have:
The original function is 
The range of this is: 
The
of the
function is the
of the
function.
<em>Hence, the domain is:</em>

Answer:
see below
Step-by-step explanation:
given:
radius of a circle = 3cm
find:
1. diameter
2. radius
3. r²
1) diameter = twice the radius = 2 r = 2 (3) = 6 cm
2) radius = 3 cm
3) r² = 3² = 9 cm