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Dahasolnce [82]

2 years ago

10

1. In what country did the government debate the merit of teaching students quadratics? 2. What modern technology would not be p

ossible without quadratics? (hint: I have one at my house.) 3. What great ideas did the Babylonians come up with? 4. Why did the Babylonians need to solve quadratic equations?

1 answer:

Alexxandr [17]2 years ago

6 0

I would say number 2 but dont take my word for it

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Any assistance would be great!

vladimir1956 [14]

The domain is the input values, which would also be X values.

{ x |x= -5, -3, 1, 2,6}

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2 years ago

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Solve for x.<br><br> 8x^2−3=2

sergeinik [125]

8x^2=2+3

8x^2=5

x^2=5/8

x= sqrt(5/8)

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2 years ago

Help!!<br> What is the relationship between the corresponding angle bisectors BD and XZ?

Harrizon [31]

Answer:

XZ = 2.5 BD

Step-by-step explanation:

Comparing the two triangles,

$\frac{BD}{BC}$ = $\frac{XZ}{XY}$

$\frac{6}{8}$ = $\frac{XZ}{20}$

⇒ XZ = $\frac{20*6}{8}$

= 15

XZ = 15

So that,

BD:XZ = 6:15

= 1:2.5

Therefore, the relationship between BD and XZ is XZ = 2.5 BD.

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2 years ago

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mrs_skeptik [129]

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2 years ago

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the half life of c14 is 5730 years. Suppose that wood found at an archeological excavation site contains about 35% as much C14 a

Furkat [3]

Answer:

The wood was cut approximately 8679 years ago.

Step-by-step explanation:

At first we assume that examination occured in 2020. The decay of radioactive isotopes are represented by the following ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$ (Eq. 1)

Where:

$\frac{dm}{dt}$ - First derivative of mass in time, measured in miligrams per year.

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

$m$ - Mass of the radioactive isotope, measured in miligrams.

Now we obtain the solution of this differential equation:

$\int {\frac{dm}{m} } = -\frac{1}{\tau}\int dt$

$\ln m = -\frac{1}{\tau} + C$

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ (Eq. 2)

Where:

$m_{o}$ - Initial mass of isotope, measured in miligrams.

$t$ - Time, measured in years.

And time is cleared within the equation:

$t = -\tau \cdot \ln \left[\frac{m(t)}{m_{o}} \right]$

Then, time constant can be found as a function of half-life:

$\tau = \frac{t_{1/2}}{\ln 2}$ (Eq. 3)

If we know that $t_{1/2} = 5730\,yr$ and $\frac{m(t)}{m_{o}} = 0.35$ , then:

$\tau = \frac{5730\,yr}{\ln 2}$

$\tau \approx 8266.643\,yr$

$t = -(8266.643\,yr)\cdot \ln 0.35$

$t \approx 8678.505\,yr$

The wood was cut approximately 8679 years ago.

5 0

2 years ago