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Gennadij [26K]
3 years ago
15

Find a quadratic polynomial whose sum and and product is -7 and -2

Mathematics
1 answer:
vampirchik [111]3 years ago
3 0

Answer:

<h2>x²+7x-2 = 0</h2>

Step-by-step explanation:

The general form of a quadratic equation with roots a and b is expressed as shown;

x²-(sum of root) x + (product of roots) = 0

x² - (a+b)x + ab = 0 ... 1

Given the sum of roots a+b = -7

Product of roots ab = -2

Substituting this values in equation 1 above wil give;

x²-(-7)x+(-2) = 0

x²+7x-2 = 0

The resulting quadratic polynomial is x²+7x-2 = 0

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–5x &lt; 35 what the answer and how to solve it
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Answer:

-7

Step-by-step explanation:

First, we know that we are solving for x so what we need to do is to find out what times -5=35.

We know that

-5x7=-35

So we know our answer cant be positive 7, which means we need the opposite of 7 which is -7.

And:

-5x-7=35

This makes the equation true:

Hence, the correct answer is -7

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2 years ago
you are mirroring the distance to your friends house you find that it is 5976 m how many kilometers is that
Effectus [21]
The distance is 5.976 kilometers
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3 years ago
Read 2 more answers
What are both of these shaded as?
Sedaia [141]

Answer:

1 whole and 15/100= <em><u>1.15</u></em>

Step-by-step explanation:

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4 0
3 years ago
Consider the following set of vectors. v1 = 0 0 0 1 , v2 = 0 0 3 1 , v3 = 0 4 3 1 , v4 = 8 4 3 1 Let v1, v2, v3, and v4 be (colu
Alona [7]

Answer:

We have the equation

c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]

Then, the augmented matrix of the system is

\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]

We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:

\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]

This matrix is in echelon form. Then, now we apply backward substitution:

1.

8c_4=0\\c_4=0

2.

4c_3+4c_4=0\\4c_3+4*0=0\\c_3=0

3.

3c_2+3c_3+3c_4=0\\3c_2+3*0+3*0=0\\c_2=0

4.

c_1+c_2+c_3+c_4=0\\c_1+0+0+0=0\\c_1=0

Then the system has unique solution that is (c_1,c_2c_3,c_4)=(0,0,0,0) and this imply that the vectors v_1,v_2,v_3,v_4 are linear independent.

4 0
3 years ago
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