Evaluate the summation below.<br> ) (-1 + 6d)<br> d=2

What is the zero in the quadratic function f(x)= 9x^2-54x-19?

Alex17521 [72]

Answer:

6.33... and 0.333...

Step-by-step explanation:

The quadratic formula is

$x=\frac{-b+\sqrt{b^2-4ac} }{2a}$ .

It is important because while some quadratics are factorable and can be solved not all are. The formula will solve all quadratic equations and can also give both real and imaginary solutions. Using the formula will require less work than finding the factors if factorable. We will substitute a=9, b=-54 and c=-19.

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}\\x=\frac{-(-54)+/-\sqrt{(-54)^2-4(9)(-19)} }{2(9)}\\x=\frac{54+/-\sqrt{2916+684} }{18}$

We will now solve for the plus and the minus.

The plus,,,

$x=\frac{54+\sqrt{3600}}{18}\\x=\frac{54+60}{18}\\x=\frac{114}{18}\\x=6.3$

and the minus...

$x=\frac{54-\sqrt{3600}}{18}\\x=\frac{54-60}{18}\\x=\frac{-6}{18}\\x=-0.333...$

8 0

3 years ago

What is the volume of the box

Sergeu [11.5K]

Answer: 240

Step-by-step explanation: the base is 40, so 4x10. volume= LxWxH, which is just all the units. 4x10x6=240. hope this helps!

6 0

2 years ago

Air pressure may be represented as a function of height (in meters) above the

denis-greek [22]

Answer: OPTION C.

Step-by-step explanation:

<h3> The complete exercise is: "Air pressure may be represented as a function of height (in meters) above the surface of the Earth, as shown below:</h3><h3> $P(h) =P_0e^{-0.00012h}$ </h3><h3> In this function $P_o$ is the air pressure at the surface of the earth, and $h$ is the height above the surface of the Earth, measured in meters. At what height will the air pressure equal 50% of the air pressure at the surface of the Earth"</h3><h3></h3>

Given the following function:

$P(h) =P_0e^{-0.00012h}$

In order to calculate at what height the air pressure will be equal 50% of the air pressure at the surface of the Earth, you can follow these steps:

1. You need to substitute $P(h)=0.5P_o$ into the function:

$0.5P_o=P_0e^{-0.00012h}$

2. Finally, you must solve for $h$ .

Remember the following property of logarithms:

$ln(b)^a=a*ln(b)\\\\ln(e)=1$

Then, you get this result:

$0.5P_o=\frac{P_o}{e^{0.00012h}}\\\\(0.5P_o)(e^{0.00012h})=P_o\\\\e^{0.00012h}=\frac{P_o}{0.5P_o}\\\\ln(e)^{0.00012h}=ln(2)\\\\0.00012h*1=ln(2)\\\\h=\frac{ln(2)}{0.00012}\\\\h=5576.2$