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MatroZZZ [7]
3 years ago
11

Given: y ll z

Mathematics
2 answers:
emmainna [20.7K]3 years ago
7 0

Answer:

As per the properties of parallel lines and interior alternate angles postulate, we can prove that:

m\angle 5+m\angle 2+m\angle 6=180^\circ

Step-by-step explanation:

<u>Given:</u>

Line y || z

i.e. y is parallel to z.

<u>To Prove:</u>

m\angle 5+m\angle 2+m\angle 6=180^\circ

<u>Solution:</u>

It is given that the lines y and z are parallel to each other.

m\angle 5, m\angle 1 are <em>interior alternate angles </em>because lines y and z are parallel and one line AC cuts them.

So, m\angle 5= m\angle 1 ..... (1)

Similarly,

m\angle 6, m\angle 3 are <em>interior alternate angles </em>because lines y and z are parallel and one line AB cuts them.

So, m\angle 6= m\angle 3 ...... (2)

Now, we know that the line y is a straight line and A is one point on it.

Sum of all the angles on one side of a line on a point is always equal to 180^\circ.

i.e.

m\angle 1+m\angle 2+m\angle 3=180^\circ

Using equations (1) and (2):

We can see that:

m\angle 5+m\angle 2+m\angle 6=180^\circ

<em>Hence proved.</em>

Finger [1]3 years ago
3 0

Answer:

Step-by-step explanation:

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This is the graph in 'slope-intercept' form. From here it is easy to see that gradient  = and that y-intercept = 490.

The easiest way to draw a straight-line graph, such as this one, is to plot the y-intercept, in this case (0, 490), then plot another point either side of it at a fair distance (for example substitute  = -5 and  = 5 to procure two more sets of co-ordinates). These can be joined up with a straight line to form a section of the graph, which would otherwise extend infinitely either side - use the specified range in the question for x-values, and do not exceed it (clearly here the limit of -values is 0 ≤ x ≤ 735, since neither x nor y can be negative within the context of the question - the upper limit was found by substituting  = 0).

In function notation, the graph is:



The graph of this function represents how the value of the function varies as the value of x varies. Looking back at the question context, this graph specifically represents how many wraps could have been sold at each number of sandwich sales, in order to maintain the same profit of $1470.

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4. Find the change in  (Δ
5. Divide the result of stage 3 by the result of stage 4
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Δ = 6 - 9 = -3
Δ = 4 - 1 = 3
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I choose the point (4, 6)
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6 = c - 4
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Within the context of the question, I imagine the prices of the two lunch specials will be the same in the third month and hence the gradient will still be  - this means steps 1-6 can be omitted. Furthermore if the axes are clearly labelled, you may even be able to just read off the y-intercept and hence dispose with steps 1-8!
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