Whats -606-(-778+20)+2n?

LESSON 3.2 (C) - Rates of Change and Slope

frozen [14]

Answer:

1. Slope = -2

2. Slope = ⁴/3

3. Slope = -3

4. Slope = 1

Step-by-step explanation:

1. Using two points on the line, (0, 0) and (-1, 2):

$slope = \frac{y_2 - y_1}{x_2 - x1} = \frac{2 - 0}{-1 - 0} = \frac{2}{-1} = -2$

2. Using two points on the line, (0, 0) and (3, 4):

$slope = \frac{y_2 - y_1}{x_2 - x1} = \frac{4 - 0}{3 - 0} = \frac{4}{3}$

3. Using two points on the line, (0, 0) and (-1, 3):

$slope = \frac{y_2 - y_1}{x_2 - x1} = \frac{3 - 0}{-1 - 0} = \frac{3}{-1} = -3$

4. Using two points on the line, (0, 0) and (2, 2):

$slope = \frac{y_2 - y_1}{x_2 - x1} = \frac{2 - 0}{2 - 0} = \frac{2}{2} = 1$

3 0

2 years ago

I= $54,p= $800 ,r = 4.5%

sattari [20]

I = PRT
I = 54
P = 800
R = 4.5%.....0.045...u have to turn it to a decimal

now we sub
54 = (800)(0.045)(T)
54 = 36T
54/36 = T
1.5 = T <=== or 1 1/2 years

7 0

3 years ago

Read 2 more answers

g You are riding a roller-coaster going around a vertical loop, on the inside of the loop. If the loop has a radius of 47.0 m, h

DaniilM [7]

Answer:

Step-by-step explanation:

Let the speed of the roller coaster at the bottom is v.

radius, r = 47 m

Let the mass is m and the force at the bottom is 2.5 times the weight.

F = 2.5 mg

So, the force is

$F = mg + \frac{mv^{2}}{r}$

$2.5 mg= mg + \frac{mv^{2}}{r}$

$1.5g=\frac{v^{2}}{r}$

v² = 1.5 gr

v² = 1.5 x 9.8 x 47

v = 26.3 m/s

7 0

3 years ago

Using a 52 card deck, how many 5 card hands have either 5 hearts or 4 hearts and 1 club

777dan777 [17]

Answer:

10,582

Step-by-step explanation:

We can choose 5 cards from 52 card deck in

$n = \binom{52}{5} = \frac{52!}{5!(52-5)!} = \frac{52!}{5!47!} = \frac{\cancel{47!} \cdot 48 \cdot 49 \cdot 50 \cdot 51 }{5! \cancel{47!}} = \frac{ 48 \cdot 49 \cdot 50 \cdot 51 }{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} = 2 \; 598 \; 960$

ways.

Now, let's calculate the number of ways we can choose 5 hearts. We know that in a 52 card deck, we have 13 hearts. Therefore, the number of ways to choose 5 hearts is

$n_1 = \binom{13}{5} = \frac{13!}{5!(13-5)!} = \frac{13!}{5!8!} = \frac{8! \cdot 9 \cdot 10 \cdot 11 \cdot 12 \cdot 13}{5!8!} = \frac{9 \cdot 10 \cdot 11 \cdot 12 \cdot 13}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} = 1287$

Similarly, number of ways to choose 4 hearts equals $\binom{13}{4}$ and number of ways to choose 1 club equals $\binom{13}{1}$ , since there are also 13 clubs in the deck.

Therefore, the number of ways of choosing 4 hearts and 1 club equals

$n_2 = \binom{13}{4} \cdot \binom{13}{1} = 9295$

The probability of this event is calculated as

$P(A) = \frac{\text{total number of ways to choose 5 hearts or 4 hearts and a club}}{\text{total number of ways to choose 5 cards from a deck of 52 cards}}$