Graphing proportional relationships
1 answer:
a. The unit rate is 9 d per t
b. Find the graph in the attachment
To graph the line, we need to find its gradient or unit rate
<h3>a. How to calculate the unit rate of change of d with respect to t?</h3>Since t has an increase of 0.20 units in t corresponds to an increase of 1.8 units in d.
The gradient or unit rate of change is m = change in d/change in t
= Δd/Δt
Since
- Δd = +1.8 units and
- Δt = +0.2 units
So, substituting the values of the variables into the equation, we have
m = Δd/Δt
= + 1.8 units/+ 0.2 units
= 9 d per t
So, the unit rate is 9 d per t
<h3>b. The graph of the line</h3>
To graph the function, we need to know the equation of the graph.
So, the equation of a graph in gradient form is
m = (y - y')/(x - x') where (x', y') = (0, 0) since the graphs both start at the origin
Since m = 9, we have
m = (y - y')/(x - x')
9 = (y - 0)/(x - 0)
9 = y/x
y = 9x
So, the equation of the line is y = 9x
Find the graph in the attachment.
Learn more about graphs here:
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Answer:
Hence proved that m∠AOD = m∠BCD.
Step-by-step explanation:
Given:
A Trapezoid with AC and BD as diagonals with O as intersection point to it.
Four angles formed at A and D are equal.
To Prove:
m∠AOD = m∠BCD
Proof:
Consider a Quadrilateral ABCD as trapezoid with diagonals as AC and BD
(Refer the Attachment).
And AB=CD which means the Quadrilateral ABCD isosceles in nature
i.e. angle A=angle D
Now
Consider triangle AOD and triangle BCD
Apply sum of all angles of triangle is 180.
And also
m∠2+∠AOD+m∠3=180
∠AOD=180-m∠2-m∠3..............equation(1)
And for Triangle BCD
m∠BCD=180-m∠3-m∠4,.....................Equation(2)
∠A=∠D................ as Quadrilateral is Isosceles.
i.e.∠2=∠3
That means by transitive property,
∠1=∠4=∠3=∠2
Using this values in above equation(1) we get ,
∠AOD=180-∠3-∠4.......................(3)
Now comparing equation 2 and 3 we get
RHS is equal for both equation that means LHS is also equal .
Therefore ∠AOD=∠BCD
Hence proved.
