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Ostrovityanka [42]
3 years ago
13

Help! Please! Geometry!!

Mathematics
1 answer:
DiKsa [7]3 years ago
5 0

9.

By the Segment Addition Postulate, SAP, we have

XY + YZ = XZ

so

YZ = XZ - XY = 5 cm - 2 cm = 3 cm

10.

M is the midpoint of XZ=5 cm so

XM = 5 cm / 2 =  2.5 cm

11.

XY + YM = XM

YM = XM - XY = 2.5 cm - 2 cm = 0.5 cm

12.

The midpoint is just the average of the coordinate A(-3,2), B(5,-4)

M = (A+B)/2 = \left( \dfrac{-3 + 5}{2}, \dfrac{2 + -4}{2} \right)

Answer: M is (1,-1)

You'll have to plot it yourself.

13.

For distances we calculate hypotenuses of a right triangle using the distnace formula or the Pythagorean Theorem.

AB^2 =(5 - -3)^2 + (-4 - 2)^2 = 8^2 + 6^2 = 100

Answer: AB=10

M is the midpoint of AB so

Answer: AM=MB=5

14.

B is the midpoint of AC.   We have A(-3,2), B(5,-4)

B = (A+C)/2

2B = A + C

C = 2B - A

C = ( 2(5) - -3, 2(-4) - 2 ) = (13, -10)

Check the midpoint of AC:

(A+C)/2 = ( (-3 + 13)/2, (2 + -10)/2 ) = (5, -4) = B, good

Answer: C is (13, -10)

Again I'll leave the plotting to you.

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General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]:                                                                     \displaystyle \lim_{x \to c} x = c

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Step-by-step explanation:

We are given the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}

When we directly plug in <em>x</em> = 0, we see that we would have an indeterminate form:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle  \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}

Plugging in <em>x</em> = 0 again, we would get:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}

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Substitute in <em>x</em> = 0 once more:

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And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

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