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3 years ago

How do you solve Binomial Squares?

1 answer:

3 0

Step 1: Square the first term of the binomial. Step 2: Multiply the first term and last term of the binomial together and then double that quantity (in other words multiply by 2). Step 3: Square the last term of the binomial

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A quadratic function is defined by f(x) = 3x^2 + 4x - 2. A linear function is defined by g(x) = mx-5. What values of the slope o

irakobra [83]

You need to find points where the line g(x) intercepts the quadratic function f(x) in one and only one point.

Then 3x^2 + 4x -2 = mx - 5

solve 3x^2 + 4x - mx -2 + 5 = 0

3x^2 + (4 - m)x + 3 = 0

In order to there be only one solution (one intersection point) the radicand of the quadratic formula must be 0 =>

b^2 - 4ac = (4 - m)^2 - 4(3)(3) = 0

(4 - m)^2 = 24

4 - m = +/- √(24)

m = 4 +/- √(24) = 4 +/- 2√(6)

Then m, the slope of the line, may be 4 + 2√6 and 4 - 2√6

7 0

3 years ago

Whats the midpoint between 8.5 and 8.6?

kaheart [24]

I would say 8.55, regardless you're talking about the median or not.

4 0

3 years ago

Read 2 more answers

When 200 gallons of oil were removed from a tank, the volume of oil left in the tank was 3/7 of the tank's capacity. what was th

ki77a [65]

Answer:

Step-by-step explanation:

"When 200 gallons of oil were removed from a tank" algebraically looks like this:

V - 200.

"...the volume of oil left in the tank was 3/7 of the tank's capacity" algebraically looks like this:

3/7(V)

Therefore, the equation is

V - 200 = 3/7(V)

Begin by multiplying both sides by 7:

7(V - 200) = 3V and

7V - 1400 = 3V so

-1400 = -4V so

V = 350 gallons

That's if the volume of the oil in the tank was 3/7 of the tank's capacity.

For the other part of the problem, we set up the equation almost the same, except the 3/7 is a 1/2:

V - 200 = 1/2(V)

Multiplying both sides by 2 gives you

2(V - 200) = V and

2V - 400 = V so

-400 = -V so

V = 400

7 0

3 years ago

Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot

Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

$\frac{dm}{dt} = -\frac{t}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in kilograms.

$t$ - Time, measured in days.

$\tau$ - Time constant, measured in days.

The solution of the differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

$\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}$

$\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }$

The half-life of a isotope ( $t_{1/2}$ ) as a function of time constant is:

$t_{1/2} = \tau \cdot \ln2$

$t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2$

The half-life difference between isotope B and isotope A is:

$\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|$

If $\frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9$ , $t_{A} = 33\,days$ and $t_{B} = 43\,days$ , the difference in the half-lives of the isotopes is: