All categories

2 years ago

Complete the following proof Given: is the midpoint of , is the midpoint of Prove: =2

Mathematics

1 answer:

Vesnalui [34]2 years ago

5 0

Answer:

Step-by-step explanation:

Statements Reasons

1). M is the midpoint of segment AB 1). Given

B is the midpoint of segment MD

2). AM = MB and MB = BD 2). Definition of midpoint

3). MD = MB + BD 3). Segment Addition Postulate

4). MD = MB + MB 4). Substitution property of of Equality

5). MD = 2MB 5). Simplify

Therefore, if M is the midpoint of segment AB, B is the midpoint of MD then MD = 2MB

You might be interested in

Use the Distributive Property to expand 4(8x - 2y)-3=

choli [55]

Answer:

32x-8y-3

Step-by-step explanation:

4(8x-2y)-3

4 x 8x = 32x

4 x 2y = -8y

5 0

2 years ago

How should they divide the money so that each class gets the same fraction of the prize money as the fraction of the box tops th

bija089 [108]

Figure the fraction of box tops each class collected, then multiply the prize money by that fraction.

Total box tops = 3760 +2301 +1855 = 7916

Mr Coronado's class's fraction: 3760/7916 × $600 = $284.99

Mrs De Souza's class's fraction: 2301/7916 × $600 = $174.41

Mr Nost's class's fraction: 1855/7916 × $600 = $140.60

5 0

3 years ago

Two dogs run around a circular track 300 m long. One dog runs at a steady rate of 15m per second, the other at a steady rate of

mihalych1998 [28]

Answer:

300 seconds

Step-by-step explanation:

The first dog run at v₁ = 15 m/sec the second one run at v₂ = 12 m/sec

we know that d = v*t then t = d/v

Then the first dog will take 300/ 12 = 25 seconds to make a turn

The second will take 300 / 15 = 20 seconds to make a turn

Then the first dog in 12 turns 12*25 will be at the start point, and so will the second one at the turn 15.

To check first dog 12 * 25 = 300

And the second dog 15 * 20 = 300

That means that time required for the two dogs to be at the start point together is

300 seconds, in that time the first dog finished the 12 turns, and the second had ended the 15.

Another procedure to solve this problem is as follows:

between 12 m/sec and 15 m/sec the minimum common multiple is 300 ( 300 is the smaller number that accept 12 and 15 as factors 12*15 = 300) Then when time arrives at 300 seconds the two dogs will be again in the starting point

3 0

2 years ago

Find a number that satifies the inequality x<-3

olga2289 [7]

It could basically be anything smaller than -3. So, some examples are -4,-5-,-6,-7,-8,-9,-10, and so on.

7 0

2 years ago

Read 2 more answers

Can you find the limits of this

Pavel [41]

Answer:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{-3}{8}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Constant]: $\displaystyle \lim_{x \to c} b = b$

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

Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

We are given the following limit:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16}$

Let's substitute in x = -2 using the limit rule:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{(-2)^3 + 8}{(-2)^4 - 16}$

Evaluating this, we arrive at an indeterminate form:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{0}{0}$

Since we have an indeterminate form, let's use L'Hopital's Rule. Differentiate both the numerator and denominator respectively:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \lim_{x \to -2} \frac{3x^2}{4x^3}$

Substitute in x = -2 using the limit rule:

$\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{3(-2)^2}{4(-2)^3}$

Evaluating this, we get:

$\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{-3}{8}$

And we have our answer.

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

Unit: Limits

6 0

2 years ago