Subtract. negative 10.0 minus left parenthesis negative 2.3 right parenthesis

If the trapezoid below is reflected across the x-axis what are the coordinates of B?

lara [203]

Answer:

answer is (2,8)

Step-by-step explanation:

b coordinate is (2,8)

8 0

2 years ago

Wonder women 1984 sold two million more than quadruple the number of tickets that were sold for Disney Soul. Disney Raya sold te

Inessa [10]

Answer:

2165

Step-by-step explanation:

add the numbers together

3 0

3 years ago

The owner of a health club decides to reduce the $35 membership fee by $5. What is the approximate percent decrease in the fee?

Aleonysh [2.5K]

Answer:

the

Step-by-step explanation:

approximate percent is 0.14%

5 0

3 years ago

Read 2 more answers

There are 4 swimmers in a race. There will be a first place, a second place, and third-place prize award. In how many different

malfutka [58]

Let's assume there's a fourth place too, and fourth place is the one where it's no prize.

4 3 2 1 (4 3 X X)
4 3 1 2 (^ but 3rd and 4th digit exchanged)
4 2 3 1 (4 2 X X)
4 2 1 3 etc.,.
4 1 2 3
4 1 3 2

As you can see, there are six possibilities assuming first swimmer was fourth.
Now, there would be six possibilities if the first swimmer was third, and same goes for second and first.
Thus, 6 x 4 = 24, there are 24 possibilities to award the 3 prizes (plus 1 no-prize)

Simple Mathematical Way:
4 x 3 x 2 x 1 = 4! = 24
There are 24 ways.

6 0

3 years ago

A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/mi

kiruha [24]

(a) after t minutes

If $y(t)$ is the amount of salt (in kg) after $t$ minutes, then $y(0) = 0 \ (pure\ water)$ and the total amount of liquid in the tank remains constant at 1000 L.

$\displaystyle \frac{dy}{dt} \\ = \left( 0.05 \frac{\text{kg}}{\text{L}} \right) \left( 5 \frac{\text{L}}{\text{min}} \right) + \left( 0.04 \frac{\text{kg}}{\text{L}} \right)\left( 10 \frac{\text{L}}{\text{min}} \right) - \left( \frac{y(t)}{1000} \frac{\text{kg}}{\text{L}} \right)\left( 15\frac{\text{L}}{\text{min}} \right) \\ \\ 
= 0.25 + 0.40 - 0.015y = 0.65 - 0.015 y = \frac{130 - 3y}{200} \frac{\text{kg}}{\text{min}}$

$\text{Hence, } \displaystyle \int \frac{dy}{130 - 3y} = \int\frac{dt}{200}$

and $-\frac{1}{3}\ln|130 - 3y| = \frac{1}{200}t + C.$

Because $y(0) = 0$ , we have $-\frac{1}{3} \ln 130 = C$

so

$-\frac{1}{3}\ln|130 - 3y| = \frac{1}{200}t -\frac{1}{3} \ln 130\ \Rightarrow \\ \\\ \ln|130 - 3y| = -\frac{3}{200}t + \ln 130 \ \Rightarrow \\ \\ |130 - 3y| = e^{-3t/200 + \ln 130} \ \Rightarrow \\ \\
|130 - 3y| = 130e^{-3t/200}$

Since $y$ is continuous, $y(0) = 0$ , and the right-hand side is never zero, we deduce that $130 - 3y$ is always positive.

Thus, $130 - y = 130e^{-3t/200}$ and
$y = \frac{130}{3}\left( 1 - e^{-3t/200}\right) \text{ kg}$

(b)

After one hour,

$y = \frac{130}{3}\left( 1 - e^{-3 \cdot 60/200}\right) = \frac{130}{3}(1 - e^{-0.9} ) \approx 25.7 \text{ kg}$

5 0

4 years ago