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Rasek [7]
3 years ago
6

Triangle ABC is isosceles.

Mathematics
2 answers:
NARA [144]3 years ago
8 0

we know that

An isosceles triangle is a triangle with two equal sides and two equal angles

so

in this problem we have

AC=AB

∠B=∠C

therefore

<u>the answer is </u>

It is equal to the measure of angle C

saw5 [17]3 years ago
8 0

Answer:  The correct option is (D) It is equal to the measure of angle C.

Step-by-step explanation:  Given that ΔABC is an isosceles triangle as shown in the figure. AB = AC.

We are to select the TRUE statement about the measure of ∠B.

We know that the measures of the angles opposite to the equal sides of a triangle are equal.

So, AB = AC implies that

m∠C = ∠B.

Therefore, the measure of angle B is equal to the measure of angle C.

Thus, (D) is the correct option.

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B is an answer.

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Anna35 [415]

Answers:

  • Part a)  \bf{\sqrt{x^2+(x^2-3)^2}
  • Part b)  3
  • Part c)   2.24
  • Part d)  1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying d = \sqrt{x^2 + (x^2-3)^2} is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

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Answer:

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Step-by-step explanation:

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Plan A cost $15 a month, including 200 free texts. After 200, they cost $0.15 each
hammer [34]
If your looking how to plot this on a graph idk but I know its something like 15.00x>200
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I think... I'm not sure... but it would make sense lol :)
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Answer:

Commutative property of multiplication

Step-by-step explanation:

Commutative property of multiplication basically means that the order in which you multiply numbers does not matter.

For exampe:

3 x 4 = 4 x 3

    12 = 12

In your case:

5 x 7 x 4 = 5 x 4 x 7

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If we break that down:

5 x 7 x 4 = 5 x 4 x 7

   35 x 4 = 20 x 7

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