- 7х — бу = 4 = – 3y + 8 ×= y=

Solve y=-3x² + 4.5x – 20 Use the Quadratic formula

statuscvo [17]

Answer:

$x=0.75 \pm 2.47067i$

Step-by-step explanation:

Quadratic Formula: $x=\frac{-b \pm \sqrt{b^2-4ac} }{2a}$

√-1 is imaginary number i

Step 1: Define

y = -3x² + 4.5x - 20

a = -3

b = 4.5

c = -20

Step 2: Substitute and Evaluate

$x=\frac{-4.5 \pm \sqrt{4.5^2-4(-3)(-20)} }{2(-3)}$

$x=\frac{-4.5 \pm \sqrt{20.25-240} }{-6}$

$x=\frac{-4.5 \pm \sqrt{-219.75} }{-6}$

$x=\frac{-4.5 \pm \sqrt{219.75}(\sqrt{-1} ) }{-6}$

$x=\frac{-4.5 \pm i\sqrt{219.75} }{-6}$

$x=\frac{4.5 \pm i14.824 }{6}$

$x=0.75 \pm 2.47067i$

7 0

3 years ago

111 is 25\%25%25, percent of what number?

vodka [1.7K]

Answer:

444

Step-by-step explanation:

111x4=444

25 percent is equal to 1/4

8 0

3 years ago

PLEASEEE I NEED HELP ASAPPPPPP DUE IN 6 MINS! PLEASE

sergey [27]

Step-by-step explanation:

60x + 3300 = 24000

60x = 24000 - 3300

60x = 20700

x = 20700 ÷ 60

x = 345

x ≤345

5 0

3 years ago

An automobile salesman receives a monthly bonus cheque if his commission income is at least $6,800.00 in a given month. There ar

Reika [66]

The salesman earns $850 per automobile he sells.

Since x represents the amount of automobiles the salesman sells, we can apply the commission as a coefficient to this variable. Therefore, the total commission that the salesman earns can be represented by $850x.

The bonus cheque is only received if the salesman's commission income is at least $6,800. 'at least' means that the salesman can still receive the cheque if his commission is exactly $6,800. The sign that we can use for this situation is the greater than or equal to sign, ≥.

The inequality that shows the commission income needed for the cheque is $850x ≥ $6,800. However, this question asks for the number of automobiles the salesman must sell to get the cheque.

Divide both sides by $850, as that represents his sales from one commission:

x ≥ 8

The inequality x ≥ 8 represents the amount of automobiles the salesman will need to sell to get the bonus cheque.

6 0

3 years ago

Find a third degree polynomial function of the lowest degree that has the zeros below and whose leading coefficient is one.

Tanzania [10]

Answer:

The polynomial function of the lowest degree that has zeroes at -1, 0 and 6 and with a leading coefficient of one is $p(x) = x^{3}-5\cdot x^{2}-6\cdot x$ .

Step-by-step explanation:

From Fundamental Theorem of Algebra, we remember that the degree of the polynomials determine the number of roots within. Since we know three roots, then the factorized form of the polynomial function with the lowest degree is:

$p(x) = (x-r_{1})\cdot (x-r_{2})\cdot (x-r_{3}) = 0$ (1)

Where $r_{1}$ , $r_{2}$ and $r_{3}$ are the roots of the polynomial.

If we know that $r_{1} = -1$ , $r_{2} = 0$ and $r_{3} = 6$ , then the polynomial function in factorized form is:

$p(x) = (x+1)\cdot x \cdot (x-6)$ (2)

And by Algebra we get the standard form of the function:

$p(x) = x\cdot (x+1)\cdot (x-6)$

$p(x) = x\cdot (x^{2}-5\cdot x -6)$

$p(x) = x^{3}-5\cdot x^{2}-6\cdot x$ (3)

The polynomial function of the lowest degree that has zeroes at -1, 0 and 6 and with a leading coefficient of one is $p(x) = x^{3}-5\cdot x^{2}-6\cdot x$ .

6 0

3 years ago

- 7х — бу = 4= – 3y + 8×=y=​

- 7х — бу = 4= – 3y + 8×=y=