Gemma wants to draw a triangle with side lengths of 4 inches, 12 inches, and 17 inches. Which statement is true? This triangle e
2 answers:
If you have a triangle with sides a, b and c, then:
In your case let a=4 in, b=12 in annd c=17 in. Check these three inequalities:
1. 4+12=16 is less than 17 - inequality a+b>c doesn't hold.
2. 4+17>12 - inequality a+c>b holds.
3. 12+17>4 - inequality b+c>a holds.
Since only two from three inequalities are true, then the triangle doesn't exist.
Answer: correct choice is D - This triangle does not exist because the sum of 4 and 12 is less than 17.
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Answer:
Hill 1: F(x) = -(x + 4)(x + 3)(x + 1)(x - 1)(x - 3)(x - 4)
Hill 2: F(x) = -(x + 4)(x + 3)(x + 1)(x - 1)(x - 3)(x - 4)
Hill 3: F(x) = 4(x - 2)(x + 5)
Step-by-step explanation:
Hill 1
You must go up and down to make a peak, so your function must cross the x-axis six times. You need six zeros.
Also, the end behaviour must have F(x) ⟶ -∞ as x ⟶ -∞ and F(x) ⟶ -∞ as x⟶ ∞. You need a negative sign in front of the binomials.
One possibility is
F(x) = -(x + 4)(x + 3)(x + 1)(x - 1)(x - 3)(x - 4)
Hill 2
Multiplying the polynomial by -½ makes the slopes shallower. You must multiply by -2 to make them steeper. Of course, flipping the hills converts them into valleys.
Adding 3 to a function shifts it up three units. To shift it three units to the right, you must subtract 3 from each value of x.
The transformed function should be
F(x) = -2(x +1)(x)(x -2)(x -3)(x - 6)(x - 7)
Hill 3
To make a shallow parabola, you must divide it by a number. The factor should be ¼, not 4.
The zeroes of your picture run from -4 to +7.
One of the zeros of your parabola is +5 (2 less than 7).
Rather than put the other zero at ½, I would put it at (2 more than -4) to make the parabola cover the picture more evenly.
The function could be
F(x) = ¼(x - 2)(x + 5).
In the image below, Hill 1 is red, Hill 2 is blue, and Hill 3 is the shallow black parabola.
The inequality to represent the temperatures at which the gas in the truck will remain in liquid form is -40 < x < 140
<h3>Part A: The inequality</h3>Let the temperature of the gas be represented with x.
When the temperature hits −40° F (or below), the gas freezes.
This means that the gas temperature must always exceed −40° F to remain being a liquid
i.e x > −40
Also, when the temperature hits 140° F (or above), the gas evaporates.
This means that the gas temperature must always be less than 140° F to remain being a liquid
i.e x < 140
So, we have:
x > −40 and x < 140
Rewrite as:
-40 < x and x < 140
Combine both inequalities
-40 < x < 140
Hence, the inequality to represent the temperatures at which the gas in the truck will remain in liquid form is -40 < x < 140
<h3>Part B: The description of the graph</h3>In (a), we have:
-40 < x < 140
The end points of the inequalities use a less than sign.
Less than signs are represented by open circle.
Hence, the graph would use an open circle, and intervals between -40 and 140 would be shaded
<h3>Part C: Would −49° F gas remained in liquid form?</h3>We have:
-40 < x < 140
The number -49 is not in the range -40 < x < 140
Hence, the gas would not remain in liquid form (in fact, it would freeze because it is less than -40)
Read more about inequalities at:
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