In these types of tables, the final box, bottom rightmost, is always the main total. This total will always add up to the same number whether you add the column or the row.
Let find it for this problem.
We can get the grand total from adding the 2 number from the row or the column.
<em>If we add column, we get, 
.</em>
<em>If we add row, we get, 
.</em>
So, our total is 166. 166 people were polled.
ANSWER: 166 People
Answer:
To isolate variables you can either:
- subtract from both sides
- add to both sides
- multiply both sides (by the same number )
- divide from both sides
These were basic.
Some more advanced ones are :
- raise both sides to the same exponent
- root both sides by the same root.
It has to be an inverse operation however, for all of these properties. And remember, they only work because they are done to both sides.
Answer:
y= 3x + 7 Step-by-step explanation:
y= -1/3x+3 (so, gradient of this line is -1/3).
formula: gradient of line one x gradient of line two= -1 (applies for 2 perpendicular lines only).
So, m1 x -1/3= -1
m1= (
)
m1= 3 (gradient of perp line)
y=mx+c
4= 3 x -1 + c
7=c
therefore, y=3x+7 (eqn of the line perp to y=-1/3x+3)
Answer:
m = 28
Step-by-step explanation:
70 × 2 = 5m
140 = 5m
28 = m
If x is a real number such that x3 + 4x = 0 then x is 0”.Let q: x is a real number such that x3 + 4x = 0 r: x is 0.i To show that statement p is true we assume that q is true and then show that r is true.Therefore let statement q be true.∴ x2 + 4x = 0 x x2 + 4 = 0⇒ x = 0 or x2+ 4 = 0However since x is real it is 0.Thus statement r is true.Therefore the given statement is true.ii To show statement p to be true by contradiction we assume that p is not true.Let x be a real number such that x3 + 4x = 0 and let x is not 0.Therefore x3 + 4x = 0 x x2+ 4 = 0 x = 0 or x2 + 4 = 0 x = 0 orx2 = – 4However x is real. Therefore x = 0 which is a contradiction since we have assumed that x is not 0.Thus the given statement p is true.iii To prove statement p to be true by contrapositive method we assume that r is false and prove that q must be false.Here r is false implies that it is required to consider the negation of statement r.This obtains the following statement.∼r: x is not 0.It can be seen that x2 + 4 will always be positive.x ≠ 0 implies that the product of any positive real number with x is not zero.Let us consider the product of x with x2 + 4.∴ x x2 + 4 ≠ 0⇒ x3 + 4x ≠ 0This shows that statement q is not true.Thus it has been proved that∼r ⇒∼qTherefore the given statement p is true.
