Ask question

Ask question

All categories

1 year ago

8

Exercise 1.2 1 Round each of these numbers to the degree of accuracy shown in brackets. a) 7765 (nearest hundred). b) 1099 (near

est thousand). c) 487 (nearest ten) d) 34 766 (nearest hundred) e) 45620000 (nearest million please help me

1 answer:

wlad13 [49]1 year ago

6 0

Answer:

a. 700

b. 1000

c. 80

d. 700

e.

You might be interested in

This is uhh due by 8 tommorow mornin please help me

Ira Lisetskai [31]

1 Simplify

1

3

x

3

1

x to

x

3

3

x

.

x

3

+

5

≤

−

4

3

x

+5≤−4

2 Subtract

5

5 from both sides.

x

3

≤

−

4

−

5

3

x

≤−4−5

3 Simplify

−

4

−

5

−4−5 to

−

9

−9.

x

3

≤

−

9

3

x

≤−9

4 Multiply both sides by

3

3.

x

≤

−

9

×

3

x≤−9×3

5 Simplify

9

×

3

9×3 to

27

27.

x

≤

−

27

x≤−27

I hope this help you

6 0

3 years ago

15% of amount = 112.79

Mazyrski [523]

To find the original amount, you divide the product by the percentage.

112.79 / 0.15 ≈ 751.933 (the 3's repeat infinitely).

To check your answer, multiply 751.933 by 0.15.
751.933 * 0.15 = 112.78995 ≈ 112.79

4 0

3 years ago

Read 2 more answers

How many squares are in this figure in all?

Tanya [424]

Answer:

explanation

Step-by-step explanation:

there

are

22

squares

all

together

6 0

3 years ago

Read 2 more answers

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

What is the equation of the line that passes through the point (2, -2) and has a<br> slope of -1/2

Naddika [18.5K]

Are you looking for this ?

3 0

3 years ago

Read 2 more answers