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3 years ago

6

Find the value of y. (27x + 4) (8x + 1) (y + 10)

1 answer:

Ostrovityanka [42]3 years ago

7 0

Ummm i need more information to help you

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In these triangles, side AB is congruent to ide DF, and see is congruent to

kompoz [17]

Answer:

hope it helps you see the attachment for further information

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3 years ago

Write an equation in slope-intercept form with a

lora16 [44]

Your equation would be:
D. y = 10x + 6

5 0

3 years ago

What two numbers are located 1/2 of a unit form 1/6 on the number line

weeeeeb [17]

Basically we need to add and subtract 1/2 from 1/6:

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

(2) 1/6 + 1/2 = 1/6 + 3/6
= 4/6
= 2/3

Therefor the two numbers that are located 1/2 unit from 1/6 are -1/3 and 2/3

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3 years ago

22. Roberto is an advertising sales representative for a magazine. He earns a commission on

liberstina [14]

Answer:

17%

Step-by-step explanation:

12,750/75,000=0.17

Which means it's 17%

Brainliest please-

6 0

3 years ago

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

3 years ago