All categories

2 years ago

Point Z(-5, -6) was translated to Z'(-8, 4). Describe the direction and distance of the

Mathematics

1 answer:

sashaice [31]2 years ago

3 0

Answer:

please send an image so I can further assist

You might be interested in

The exponential model A = 661.7 e^0.011t describes the population, A, of a country in millions, t years after 2003. Use the mode

faust18 [17]

Answer:

661.7 million

Step-by-step explanation:

Given the exponential model :

A = 661.7 e^0.011t

The general form of an Exponential model is expresses as :

A = A0 * e^rt

Where A = final value ; A0 = Initial value ; r = growth rate and t = time elapsed

From the question t = time after 2003

Therefore, A0 = initial population, which is the population in 2003

Therefore, A0 = 661.7

Or we could put t = 0 in the equation and solve for A

A = 661.7 e^0.011(0)

A = 661.7 * 1

A = 661.7

Hence, population in 2003 is 661.7 million

7 0

2 years ago

Read the statement.

Sladkaya [172]

Answer:

Answer: Option (A) is true "p → q represents the original conditional

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Which is the correct stem-and-leaf plot for the data set?<br> 11, 10, 49, 36, 35, 34, 11, 32

hram777 [196]

Answer:

The answer is B

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

2 years ago

Find the area of the shape below.

Otrada [13]

Area = 10.08 yd²

Solution:

The given shape is a trapezoid.

Length of top base ( $b_1$ ) = 5.4 yd

Length of bottom base ( $b_2$ ) = 1.8 yd

Height of trapezoid (h) = 2.8 yd

<u>To find the area of the trapezoid:</u>

Area of the trapezoid = $\frac{1}{2}\times (b_1+b_2)\times h$

$$=\frac{1}{2}\times (5.4+1.8)\times 2.8$

$$=\frac{1}{2}\times 7.2\times 2.8$

= 10.08 yd²

Area = 10.08 yd²

3 0

2 years ago