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3 years ago

Q # 3 evaluate the polynomial

Mathematics

2 answers:

Bad White [126]3 years ago

6 0

Heres what you do: Replace X with -3 and Y with 2. Your equation will look like this: 6(-3)-2. Multiply 6 and -3 to get -18. Then subtract 2 from -18. Your answer is -20

Zielflug [23.3K]3 years ago

5 0

For this case we have the following polynomial:
$6x - y
$
We observe that we have a polynomial of two variables.
The values of the variables are given by:
$x = -3

y = 2$
Therefore, substituting values we have:
$6 (-3) - (2)
$
Rewriting we have:
$-18 - 2 = -20
$
Answer:
The polynomial evaluated for the given values results in:
-20
option 2

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Please help!! What is the solution to the quadratic inequality? 6x2≥10+11x

fredd [130]

Answer:

The solution of the inequation $6\cdot x^{2} \geq 10 + 11\cdot x$ is $\left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right)$ .

Step-by-step explanation:

First of all, let simplify and factorize the resulting polynomial:

$6\cdot x^{2} \geq 10 + 11\cdot x$

$6\cdot x^{2}-11\cdot x -10 \geq 0$

$6\cdot \left(x^{2}-\frac{11}{6}\cdot x -\frac{10}{6} \right)\geq 0$

Roots are found by Quadratic Formula:

$r_{1,2} = \frac{\left[-\left(-\frac{11}{6}\right)\pm \sqrt{\left(-\frac{11}{6} \right)^{2}-4\cdot (1)\cdot \left(-\frac{10}{6} \right)} \right]}{2\cdot (1)}$

$r_{1} = \frac{5}{2}$ and $r_{2} = -\frac{2}{3}$

Then, the factorized form of the inequation is:

$6\cdot \left(x-\frac{5}{2}\right)\cdot \left(x+\frac{2}{3} \right)\geq 0$

By Real Algebra, there are two condition that fulfill the inequation:

a) $x-\frac{5}{2} \geq 0 \,\wedge\,x+\frac{2}{3}\geq 0$

$x \geq \frac{5}{2}\,\wedge\,x \geq-\frac{2}{3}$

$x \geq \frac{5}{2}$

b) $x-\frac{5}{2} \leq 0 \,\wedge\,x+\frac{2}{3}\leq 0$

$x \leq \frac{5}{2}\,\wedge\,x\leq-\frac{2}{3}$

$x\leq -\frac{2}{3}$

The solution of the inequation $6\cdot x^{2} \geq 10 + 11\cdot x$ is $\left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right)$ .

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Answer:

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Step-by-step explanation:

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