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3 years ago

Find the slope between (2,-3) and (4,3)

Mathematics

1 answer:

cupoosta [38]3 years ago

8 0

Answer:

Step-by-step explanation:

var y = 6

var x = 2

6/2 = 3

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The points (-1.6, 2.7) and (3.2, 7.98) are on the graph of a linear relationship between two variables, r and y.

Aleksandr [31]

Answer:

y = 1.1x +4.46

y = 129.86 for x = 114

Step-by-step explanation:

The two-point form of the the equation for a line is useful for this.

y = (y2 -y1)/(x2 -x1)(x -x1) +y1

y = (7.98 -2.7)/(3.2 -(-1.6))(x -(-1.6)) + 2.7

y = 5.28/4.8(x +1.6) + 2.7

y = 1.1x +1.76 +2.7

y = 1.1x +4.46

When x=114, the value of y is ...

y = 1.1(114) +4.46

y = 129.86

6 0

3 years ago

If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by

Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by $y = 70t-16t^2$ .

To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

$(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}$

$(\text{Average velocity})_{0.1\ s}=696\ ft/s$

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

$(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}$

$(\text{Average velocity})_{0.01\ s}=7536\ ft/s$

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

$(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}$

$(\text{Average velocity})_{0.001\ s}=75936\ ft/s$

5 0

4 years ago

A football player lost 10 yards when he was tackled. a 10 b -10

Ganezh [65]

The key word “lost” indicates that the football player went back 10 yards. The correct answer is B. -10! :)

3 0

3 years ago

Read 2 more answers

What's the answer to this math problem?

m_a_m_a [10]

Answer:

take the numbers and do the equation steps then fill the box in

Step-by-step explanation:

8 0

3 years ago

Marlena was asked to find an expression that is not equivalent to 212 Which of the following is not equivalent to the given

True [87]

Answer:

(2⁶)(2⁶)

Step-by-step explanation:

Hope it help

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7 0

3 years ago