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SOVA2 [1]
2 years ago
5

In GeoGebra, display the slope of AB and the slope of the perpendicular line passing through C. Use this to verify your response

s in parts B and C. Then move points A, B, and C on the grid to several different locations, and record the slopes of the two lines and the coordinates of A, B, and C.
Mathematics
1 answer:
kati45 [8]2 years ago
5 0

The slope of the perpendicular line passes through C in GeoGebra.

<h3>What is GeoGebra?</h3>

GeoGebra is the software for the mathematical question for all levels of academics.

GeoGebra brings many mathematical problem solutions like geometry, algebra, spreadsheets, graphing, statistics, and calculus in one search engine.

Display the slope of and the slope of the perpendicular line passing through C.

Hence, the slope of the perpendicular line passes through C in GeoGebra.

For more details about GeoGebra:

brainly.com/question/10400398

#SPJ1

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Orange M&amp;M’s: The M&amp;M’s web site says that 20% of milk chocolate M&amp;M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

                 =0.1968

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0
3 years ago
Prove that:<br>tan80°=2tan 70°+ tan10°​
Angelina_Jolie [31]
Tan80=5.67
2tan70+tan10=5.67

So tan80=2tan70+tan10
8 0
3 years ago
Read 2 more answers
Solve for v -2(3v-3)+2v=8+2(3v-9)
sweet [91]

Answer: V=8/5 or V=1.6 or V=1 ,3/5  / is over i dont know how the write 8

                                                                                                                            --

                                                                                                                             5

on here

5 0
4 years ago
Read 2 more answers
Help!!!! 20 points!!!!!!’
Gnoma [55]

Answer:

option c is the correct answer

first take y raised to the power 4 common then cancel its power by y raised to the power 3

you get your answer

3 0
3 years ago
What are the roots of the polynomial equation x^3-5x+5=2x^2-5? Use a graphing calculator and a system of equations. Round nonint
leonid [27]

The right answer is c. –2.24, 2, 2.24


This question needs to be solved in two ways. First, using a graphing calculator. Next, using a system of equations.


1. Using a graphing calculator.


We have the following polynomial equation:

x^3-5x+5=2x^2-5


By ordering this equation we have:

x^3-2x^2-5x+10=0


So, we can say that this equation comes from a function given by:

f(x)=x^3-2x^2-5x+10


Thus, by plotting this function, we have that the graph of this function is indicated in Figure 1. By zooming, we can see, in Figure 2, that the roots of the polynomial equation are the x-intercepts of f(x) which are:


x_{1}=-2.236 \\ \\ x_{2}=2 \\ \\ x_{3}=2.236


Finally, rounding noninteger roots to the nearest hundredth we have:


\boxed{Root_{1}=-2.24} \\ \\ \boxed{Root_{2}=2} \\ \\ \boxed{Root_{3}=2.24}


2. Using a system of equations.


The ordered equation is:

x^3-2x^2-5x+10=0


By arranging to factor out we have:

x^3-5x-2x^2+10=0


Then, by factoring:

x(x^2-5)-2(x^2-5)=0


Term (x^2-5) is a common factor, thus:


(x-2)(x^2-5)=0 \\ \\ (x-2)(x-\sqrt{5})(x+\sqrt{5})=0 \\ \\ Finally: \\ \\ \boxed{Root_{1}=-\sqrt{5}=-2.24} \\ \\ \boxed{Root_{2}=2} \\ \\ \boxed{Root_{3}=\sqrt{5}=2.24}

6 0
4 years ago
Read 2 more answers
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