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Gnesinka [82]
3 years ago
9

1st drop down answers are

Mathematics
2 answers:
bixtya [17]3 years ago
7 0
It has to be not valid if you were choose other you would get wrong
faust18 [17]3 years ago
5 0

Answer:

uhhh i think all postive.

Step-by-step explanation:

conclusion drawings

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What is the sum of the roots of the polynomial shown below?<br> f(x) = x^3 + 2x^2 - 11x-12
svp [43]

<u>Answer:  </u>

Sum of the roots of the polynomial x^{3}+2 x^{2}-11 x-12 \text { is }-2

<u>Solution:</u>

The general form of cubic polynomial is a x^{3}+b x^{2}+c x+d=0 ---- (1)

If we have any cubic polynomial a x^{3}+b x^{2}+c x+d=0 having roots \alpha , \beta , \theta

Sum of roots \alpha + \beta + \theta = \frac{-b}{a} ---(2)

From question given that,

x^{3}+2 x^{2}-11 x-12 --- (3)

On comparing equation (1) and (3), we get a = 1, b = 2, c = -11 and d = -12

Hence the sum of roots using eqn 2 is given as,

= \frac{-2}{1}

= -2

Hence the sum of the roots of the polynomial x^{3}+2 x^{2}-11 x-12 \text { is }-2

4 0
3 years ago
Can someone tell me how many cds in both columns combined be?<br>​
Keith_Richards [23]

Answer:

2021= 700cds  2020=750cds

Step-by-step explanation:

Puedes ver como indica en el lado izquierdo que la barra de cds de el 2021 esta en la mitad de 800 y 600 que es igual a 700 cds en 2021

Y en la parte del 2020 se ve que este en la mitad de 800 y 700 considerando el calculo anterior por lo que daría 750 cds en 2020

6 0
1 year ago
Every workday, Zoe bikes 2.6 miles to work along city streets. On the way back, she likes to take a shortcut through a park, so
snow_lady [41]

Answer:

I'm pretty sure it's 7.1 miles in a week

8 0
3 years ago
Read 2 more answers
Let V=ℝ2 and let H be the subset of V of all points on the line 4x+3y=12. Is H a subspace of the vector space V?
sergejj [24]

Answer:

No, it isn't.

Step-by-step explanation:

We have V=IR^{2} and let H be the subset of V of all points on the line

4x+3y=12

We need to find if H is a subspace of the vector space V.

In IR^{2} all the possibilities for own subspace of the vector space IR^{2} are :

  • IR^{2} itself.
  • The vector 0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]
  • All lines in IR^{2} that passes through the origin  (  0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]  )

We know that H is the subset of IR^{2} of all points on the line 4x+3y=12

If we look at the equation, the point \left[\begin{array}{c}0&0\end{array}\right] doesn't verify it because :

4x+3y=12\\4(0)+3(0)=12\\0=12

Which is an absurd. Therefore, H doesn't contain the origin (and H is a line in IR^{2}). Finally, it can't be a vector space of V=IR^{2}

8 0
3 years ago
Please help asap 25 pts
Elenna [48]

The answer is A.  Those are the four points that are graphed.

3 0
3 years ago
Read 2 more answers
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