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3 years ago

Can someone please help me on this one?

Mathematics

2 answers:

Hunter-Best [27]3 years ago

7 0

The one on the bottom right surface

siniylev [52]3 years ago

7 0

Answer is bottom left
x y
1 4
2 8
3 12

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3 years ago

Construct a polynomial function with the following properties: third degree, only real coefficients, −3 and 3+i are two of the z

irina [24]

Answer:

$\boxed{-3(x+3)(x^2-6x+10)}$

Step-by-step explanation:

Hello,

As the polynomial has only real coefficients, it means that 3-i is a zero too, because we apply the Conjugate Zeros Theorem.

It means that we can write the expression as below, k being a real number that we will have to identify.

$k(x+3)(x-3-i)(x-3+i)=k(x+3)((x-3)^2-i^2)\\\\=k(x+3)(x^2-6x+9+1)\\\\=k(x+3)(x^2-6x+10)$

And for x = 0, y = -90 so we can write

-90=k*3*10, meaning that k=-3

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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3 years ago

Find y Solve the simultaneous equations 3x + 2y = 27 3x - Y = 9

lutik1710 [3]

Answer:

$y = 6$

Step-by-step explanation:

To find $y$ , we need to eliminate $x$ in this system of equations:

$3\cdot x + 2\cdot y = 27$ (1)

$3\cdot x - y = 9$ (2)

From (1) and (2):

$x = \frac{27-2\cdot y}{3}$

$x = \frac{9+y}{3}$

Then, we equalize both expressions and solve for $y$ :

$\frac{27-2\cdot y}{3} = \frac{9+y}{3}$

$27-2\cdot y = 9 + y$

$3\cdot y = 18$

$y = 6$

6 0

3 years ago