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lakkis [162]
2 years ago
15

Help plssss! i’m rly struggling in geometry

Mathematics
1 answer:
Lelechka [254]2 years ago
7 0

Answer:

i don't know

Step-by-step explanation:

fffgg

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Suppose the roots of the polynomial $x^2 - mx + n$ are positive prime integers (not necessarily distinct). Given that $m < 20
Vsevolod [243]

Answer:

<em>18</em> values for n are possible.

Step-by-step explanation:

Given the quadratic polynomial:

$x^2 - mx + n$

such that:

Roots are positive prime integers and

$m < 20$

To find:

How many possible values of n are there ?

Solution:

First of all, let us have a look at the sum and product of a quadratic equation.

If the quadratic equation is:

Ax^{2} +Bx+C

and the roots are: \alpha and \beta

Then sum of roots, \alpha+\beta = -\frac{B}{A}

Product of roots, \alpha \beta = \frac{C}{A}

Comparing the given equation with standard equation, we get:

A = 1, B = -m and C = n

Sum of roots,  \alpha+\beta = -\frac{-m}{1} = m

Product of roots, \alpha \beta = \frac{n}{1} = n

We are given that m  

\alpha and \beta are positive prime integers such that their sum is less than 20.

Let us have a look at some of the positive prime integers:

2, 3, 5, 7, 11, 13, 17, 23, 29, .....

Now, we have to choose two such prime integers from above list such that their sum is less than 20 and the roots can be repetitive as well.

So, possible combinations and possible value of n (= \alpha \times \beta) are:

1.\ 2,  2\Rightarrow  n = 2\times 2 = 4\\2.\ 2, 3 \Rightarrow  n = 6\\3.\ 2, 5 \Rightarrow  n = 10\\4.\ 2,  7\Rightarrow  n = 14\\5.\ 2, 11 \Rightarrow  n = 22\\6.\ 2, 13 \Rightarrow  n = 26\\7.\ 2, 17 \Rightarrow  n = 34\\8.\ 3,  3\Rightarrow  n = 3\times 3 = 9\\9.\ 3, 5 \Rightarrow  n = 15\\10.\ 3, 7 \Rightarrow  n = 21\\

11.\ 3,  11\Rightarrow  n = 33\\12.\ 3, 13 \Rightarrow  n = 39\\13.\ 5, 5 \Rightarrow  n = 25\\14.\ 5, 7 \Rightarrow  n = 35\\15.\ 5, 11 \Rightarrow  n = 55\\16.\ 5, 13 \Rightarrow  n = 65\\17.\ 7, 7 \Rightarrow  n = 49\\18.\ 7, 11 \Rightarrow  n = 77

So,as shown above <em>18 values for n are possible.</em>

3 0
3 years ago
X^2-x+12 solve using the quadratic formula
devlian [24]

The quadratic formula is x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}.

From the problem, a is 1, b is -1, and c is 12. Plugging in these values gives:

x=\frac{-(-1(-1)\pm\sqrt{(-1)^2-4(1)(12)} }{2(1)}

Because there is a square root of a negative number, there is no solution.

4 0
3 years ago
Which is equivalent to the complex number i^32?
Ksivusya [100]

Answer:

where i^2 = -1 (minus one)

So, i^2 = -1

i^4 = (i^2)^2 = (-1)^2 = -1 X-1 = 1

like ;

i^32 = (i^2)^16 = (-1)^16 =  -1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X

     = 1

Step-by-step explanation:

4 0
3 years ago
IMPORTANT PLEASE!! 10 POINTS :)
s344n2d4d5 [400]

Answer:

The domain is all the possible inputs for this function. The range is all the possible outputs for this function.

The line starts at the <em>x-value</em> of 0 and ends at the <em>x-value</em> of 12. Therefore, your domain is D: [0, 12].

The line starts at the <em>y-value</em> of 9000 and ends at the <em>y-value</em> of 0. Therefore, your range is R: [0, 9000].

5 0
3 years ago
In your path class, 60% of the student are girls. If there are 15 girls in the class, how many students are in your math class?
xxMikexx [17]
25 students in class
3 0
3 years ago
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