Help with finding the slope and graphing just the questions in general

Mathematics

1 answer:

irinina [24]2 years ago

4 0

Start at (0,84) as your y-intercept because thats what he starts with. Then make a negative slope decreasing by 2 each time the x value is increased by 1

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Mario’s company makes unusually shaped imitation gemstones. One gemstone has 14 faces and 10 vertices. How many edges does the g

Ivan

The gemstone has 22 edges.

We use Euler's formula on this:

F + V - E = 2

Substituting our information, we have:

14 + 10 - E = 2

Combining like terms,
24 - E = 2

Subtract 24 from both sides:
24 - E - 24 = 2 - 24
-E = -22

Divide both sides by -1:
-E/-1 = -22/-1
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Answer:

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Step-by-step explanation:

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Looking at the two quadratic functions below (1 & 2), answer the following questions.

algol [13]

Part A:

Given parabola (1) to be $f(x) =-(x+12)^2 -6$ , and parabola (2) to be $f(x)=13(x-4)^2+1$

Notice that parabola (2) is stretched horizontally by a factor of 13 which is greater than 1. This means that parabola (2) is further away from the x-axis than parabola (1). (i.e. parabola (2) is more 'vertical' than parabola (1).

Therefore, parabola (1) is wider than parabola (2).

Part B:

A parabola open up when the coefficient of the quadratic term (the squared term) is positive and opens down when the coefficient of the quadratic term is negative.

Given parabola (1) to be $f(x) =-(x+12)^2 -6$ , and parabola (2) to be $f(x)=13(x-4)^2+1$

Notice that the coefficient of the quadratic term is positive for parabola (2) and negative for parabola (1), therefore, the parabola that will open down is parabola (1).

Part C:

For any function, f(x), the graph of the function is moved p places to the left when p is added to x (i.e. f(x + p)) and moves p places to the right when p is subtracted from x (i.e. f(x - p)).

Given parabola (1) to be $f(x) =-(x+12)^2 -6$ , and parabola (2) to be $f(x)=13(x-4)^2+1$

Notice that in parabola (1), 12 is added to x, which means that the graph of the parent function is shifted 12 places to the left while in parabola (2), 4 is subtracted from x, which means that the graph of the parent function is shifted 4 places to the right.

Therefore, the parabola that would be furthest left on the x-axis is parabola (1).

Part D:

For any function, f(x), the graph of the function is moved q places up when q is added to the function (i.e. f(x) + q) and moves q places down when q is subtracted from the function (i.e. f(x) - q).

Given parabola (1) to be $f(x) =-(x+12)^2 -6$ , and parabola (2) to be $f(x)=13(x-4)^2+1$

Notice that in parabola (2), 1 is added to the function, which means that the graph of the parent function is shifted 1 place up while in parabola (1), 6 is subtracted from the function, which means that the graph of the parent function is shifted 6 places down.

Therefore, the parabola that would be highest on the y-axis is parabola (2).

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zvonat [6]

Answer:

to answer questions according to how it is

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