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ryzh [129]
2 years ago
14

Determine the discriminant for the quadratic equation 0 = –2x2 + 3. Based on the discriminant value, how many real number soluti

ons does the equation have?
Discriminant = b2 – 4ac
Mathematics
2 answers:
nydimaria [60]2 years ago
8 0
Hello,

-2x²+3=0
Δ=0²+4*2*3=24

==>2 reals solutions

Sedaia [141]2 years ago
8 0

Answer:

2 is the real solution

Step-by-step explanation:

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Mrs.Steffen’s third grade class has 30 students in it. The students are divided into three groups(numbered 1, 2,and 3),each havin
qaws [65]

Answer:

a. \\ 10! = 3628800;

b. \\ 10!*10!*10! = 47784725839872000000 = 4.7784725839872*10^{19}

Step-by-step explanation:

We need here to apply the <em>Multiplication Principle </em>or the <em>Fundamental Principle of Counting</em> for each answer. Answer <em>b</em> needs an extra reasoning for being completed.

The <em>Multiplication Principle</em> states that if there are <em>n</em> ways of doing something and <em>m</em> ways of doing another thing, then there are <em>n</em> x <em>m</em> ways of doing both (<em>Rule of product</em> (2020), in Wikipedia).

<h3>In how many ways can ten students line up? </h3>

There are <em>ten</em> students. When one is selected, there is no other way to select it again. So, <em>no repetition</em> is allowed.

Then, in the beginning, there are 10 possibilities for 10 students; when one is selected, there are nine possibilities left. When another is selected, eight possibilities are left to form the file, and so on.

Thus, we need to multiply the possibilities after each selection: that is <em>why</em> the <em>Multiplication Principle</em> is important here.

This could be expressed mathematically using n!:

\\ n! = n * (n-1)! * (n-2)! *...* 2*1.

For instance, \\ 5! = 5 * (5-1)! * (5-2)! *...*2*1 = 5 * 4 * 3 * 2 * 1 = 120.

So, for the case in question, the <em>ten</em> students can line up in:

\\ 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3628800 ways to line up in a single file.

<h3>Second Question</h3>

For this question, we need to consider the former reasoning with extra consideration in mind.

The members of Group 1 can occupy <em>only</em> the following places in forming the file:

\\ G1 = \{ 1, 4, 7, 10, 13, 16, 19, 22, 25, 28\}^{th} <em>places</em>.

The members of Group 2 <em>only</em>:

\\ G2 = \{ 2, 5, 8, 11, 14, 17, 20, 23, 26, 29\}^{th} <em>places</em>.

And the members of Group 3, the following <em>only</em> ones:

\\ G3 = \{ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30\}^{th} <em>places.</em>

Well, having into account these possible places for each member of G1, G2 and G3, there are: <em>10! ways</em> for lining up members of G1; <em>10! ways</em> for lining up members of G2 and, also, <em>10! ways</em> for lining up members of G3.

After using the <em>Multiplication Principle</em>, we have, thus:

\\ 10! * 10! * 10! = 47784725839872000000 = 4.7784725839872 *10^{19} <em>ways the students can line up to come in from recess</em>.

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2 years ago
What is the word form of 34/100
valkas [14]
34/100
= 0.34

3 tenths, 4 hundreths
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Ms. Velez will use both x gray bricks and y red bricks to build a wall around her garden. Gray bricks cost $0.45 each and red br
Alekssandra [29.7K]

It is given in the question that

Ms. Velez will use both x gray bricks and y red bricks to build a wall around her garden. Gray bricks cost $0.45 each and red bricks cost $0.58 each. She can spend up to $200 on her project, and wants the number of red bricks to be less than half the number of gray bricks.

Maximum she can spend is $200. That is

0.45x + 0.58y \leq 200&#10;\\  y< \frac{1}{2} x

And

x\geq 0 , y \geq 0

And that's the required inequalities .

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Three friends went running. The table below shows their last 3 lap times.
Goryan [66]

Answer: 1 min 50 sec

Step-by-step explanation:

The range is max value - min value

min value = 14 mins 40 sec

max value = 16 mins 30 sec

so now subtract these 16 mins 30 sec - 14 mins 40 sec

= 1 min 50 sec

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If i know a real root of f(x) = x3 -6x2 + 11x – 6 is neither 1, even, or negative, a good guess would be?
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