Answer:
The man has 6 GHc20.00 notes and 4 GHc10.00 left
Step-by-step explanation:
Given
Amount = 3 GH¢50.00, 7 GH¢20.00 and 5 GH¢10.00
Spendings = GH¢80,00
Required
Determine how many GH¢20.00 and GH¢10.00 notes left
The amount he has can be represented as:
This can be represented as:
Represent:
GHc50.00 with x, GHc20.00 with y and GHc10.00 with z
So, we have:
Subtract the spendings from the total to get amount left
Collect Like Terms
Recall that, we have represented:
GHc50.00 with x, GHc20.00 with y and GHc10.00 with z
<em>This means that, the man has 6 GHc20.00 notes and 4 GHc10.00 left</em>
Answer:
Step-by-step explanation:
we have
----> equation A
----> equation B
Multiply the equation B by 2 both sides
------> equation C
Adds equation A and equation C
Find the value of y
the solution is the point
Answer:
The values are evaluated below.
Step-by-step explanation:
Given function is
We have to find the domain, range, rel max, rel min, end behaviour, increasing or decreasing intervals and zeros of polynomial.
Domain:
The domain of a function is the set of all possible values of x for the given function.
Here domain is set of all real numbers R.
Range:
The range is the resulting y-values we get after substituting all the possible x-values.From the graph we see that
The range is 
Relative maxima and minima of a function, are the largest and smallest value of the function on an entire domain of function.
Relative max of is 0 at x=0
and 
at 
Relative min is
at 
From the graph we see that
End behaviour is As 
Increasing intervals and decreasing intervals are
Increasing: 
Decreasing: 
Zeroes are :
⇒ 
⇒
Hence, zeroes are 0, 0, -3, -4
Area = π x radius²
dA/dr = 2πr
Average rate of change = (rate of change at one instant + rate of change at other instant)/2
= (2πr₁ + 2πr₂)/2
= (r₁ + r₂)π
Ai) 9π
Aii) 8.5π
Aiii) 8.1π
B) Instantaneous change:
(dA/dr) at (r = 4) = 2π(4)
= 8π
