What is the value of b^2 - 4ac for the following equation? 2x^2 + 3x = -1

PLEASE ANSWER! Given the functions f(x) = x2 + 6x - 1, g(x) = -x2 + 2, and h(x) = 2x2 - 4x + 3, rank them from least to greatest

Ratling [72]

Answer:

he rank from least to great based on their axis of symmetry:

0, 1, -3 ⇒ g(x), h(x), f(x)

So, option C is correct.

Step-by-step explanation:

A quadratic equation is given by:

$ax^2+bx+c =0$

Here, a, b and c are termed as coefficients and x being the variable.

Axis of symmetry can be obtained using the formula

$x = \frac{-b}{2a}$

Identification of a, b and c in f(x), g(x) and h(x) can be obtained as follows:

$f(x) = x^2 + 6x - 1$

⇒ a = 1, b = 6 and c = -1

$g(x) = -x^2 + 2$

⇒ a = -1, b = 0 and c = 2

$h(x) = 2^2 - 4x + 3$

⇒ a = 2, b = -4 and c = 3

So, axis of symmetry in $f(x) = x^2 + 6x - 1$ will be:

$x = \frac{-b}{2a}$

x = -6/2(1) = -3

and axis of symmetry in $g(x) = -x^2 + 2$ will be:

$x = \frac{-b}{2a}$

x = -(0)/2(-1) = 0

and axis of symmetry in $h(x) = 2^2 - 4x + 3$ will be:

$x = \frac{-b}{2a}$

x = -(-4)/2(2) = 1

So, the rank from least to great based on their axis of symmetry:

0, 1, -3 ⇒ g(x), h(x), f(x)

So, option C is correct.

Keywords: axis of symmetry, functions

Learn more about axis of symmetry from brainly.com/question/11800108

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7 0

3 years ago

Would appreciate some help with this: Verify (sinq+cosq)^2 - 1 = sin2q

Phantasy [73]

Hello from MrBillDoesMath!

Answer:

See Discussion below

Discussion:

(sinq + cosq)^2 = => (a +b)^2 = a^2 + 2ab + b^2

(sinq)^2 + (cosq)^2 + 2 sinq* cosq => as (sinx)^2 + (cosx)^2 = 1

1 + 2 sinq*cosq (*)

Setting a = b = q in the trig identity:

sin(a+b) = sina*cosb + cosa*sinb

sin(2q) = (**)

sinq*cosq + cosq*sinq => as both terms are identical

2 sinq*cosq

Combining (*) and (**)

(sinq + cosq)^2 = 1 + 2sinq*cosq => (**) 2sinq*cosq = sqin(2q)

= 1 + sin(2q)

Hence

(sinq + cosq)^2 = 1 + sin(2q) => subtracting 1 from both sides

(sinq + cosq)^2 - 1 = sin(2q)

The last statement is what we are trying to prove.

Thank you,

MrB

7 0

3 years ago

The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%

liq [111]

<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by $\frac{100+1.8}{100}$ = $\frac{101.8}{100}$ .

So, the population in the year t can be given by $P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}$

Population in the year 2000 = $3,381,000\textrm{x}(\frac{101.8}{100})^{6}$ = $3,762,979.38$

Population in year 2000 = 3,762,979

Let us assume population doubles by year $y$ .

$2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}$

$log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})$

$y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537$

$y$ ≈ $2033$

∴ By 2033, the population doubles.

4 0

3 years ago

Helena has saved $595 to put toward snowboarding equipment and lift tickets at the hill. Each lift ticket costs $35. If she want

Akimi4 [234]

595 - 35t > 420
-35t > 420 - 595
-35t > -175
t < 5........so she can buy 4 tickets

5 0

3 years ago

Why do we need to integrate probabilities in statistics?

geniusboy [140]

For the answer to the question above asking why do we need to integrate probabilities in statistics?
Well, integration is used very very often in theoretical statistics. Transformation relates to the result that data values that are modelled as being random variables from any given continuous distribution can be converted to random variables having a standard uniform distribution. So, we need to see other possibilities by combining (one thing) with another.

5 0

3 years ago