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3 years ago

What is the distance between the following points? WILL GIVE BRAINLIEST!!

Mathematics

1 answer:

nexus9112 [7]3 years ago

8 0

Answer:

A. 5

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS 

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Reading a coordinate plane
Coordinates (x, y)

Algebra II

Distance Formula: $\displaystyle d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Step-by-step explanation:

Step 1: Define

Identify points from graph.

Point (8, 5)

Point (4, 2)

Step 2: Find distance d

Simply plug in the 2 coordinates into the distance formula to find distance d

Substitute in points [Distance Formula]: $\displaystyle d = \sqrt{(8 - 4)^2 + (5 - 2)^2}$
[√Radical] (Parenthesis) Subtract: $\displaystyle d = \sqrt{4^2 + 3^2}$
[√Radical] Evaluate exponents: $\displaystyle d = \sqrt{16 + 9}$
[√Radical] Add: $\displaystyle d = \sqrt{25}$
[√Radical] Evaluate: $\displaystyle d = 5$

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3 years ago

Read 2 more answers

If a ball is thrown in the air with a velocity 52 ft/s, its height in feet t seconds later is given by y = 52t - 16t^2. Find the

Sunny_sXe [5.5K]

Answer:

a.) -20ft/s

b.) -13.6ft/s

c.) -12.8ft/s

d.) -12.16ft/s

e.) -12ft/s

Step-by-step explanation:

Average Velocity = Change in distance/change in time.

Distance from the question in given in form of t as y= 52t - 16t² If our initial time is 2, distance travelled at t=2 is given as

Y(2) = 52(2) - 16(2)² =104 - 64

Y = 40ft.

For question a, when change In t is 0.5 seconds, that is from 2 sec to 2.5 seconds,

Average velocity = y(2.5) - y(2)/0.5

y(2.5) = 52(2.5) - 16(2.5)² =130 - 100 = 30

Y(2) = 40

Average velocity in 0.5 seconds = [30 - 40]/0.5 = -20ft/s.

For question b, when change In t is 0.1 seconds, that is from 2 sec to 2.1 seconds,

Average velocity = y(2.1) - y(2)/0.1

y(2.1) = 52(2.1) - 16(2.1)² =109.2 - 70.56= 38.64

Y(2) = 40,

Average velocity in 0.1 seconds = [38.64 - 40]/0.1 = -13.6ft/s.

For question c, when change In t is 0.05 seconds, that is from 2 sec to 2.05 seconds,

Average velocity = y(2.05) - y(2)/0.05

y(2.05) = 52(2.05) - 16(2.05)² = 106.6 - 67.24 = 39.36

Y(2) = 40

Average velocity in 0.05 seconds = [39.36 - 40]/0.05 = -12.8ft/s.

For question d, when change In t is 0.01 seconds, that is from 2 sec to 2.01 seconds,

Average velocity = y(2.01) - y(2)/0.01

y(2.01) = 52(2.01) - 16(2.01)² = 104.52 -64.6416 = 39.8784

Y(2) = 40

Average velocity in 0.5 seconds = [39.8784 - 40]/0.01 = -12.16ft/s.

Instantaneous velocity at t =2 is derived by getting the first derivative of y and inserting Our value of t=2 into the first derivative.

If y = 52t - 16t², then derivative of y becomes y' given as

y'= 52 - 32t

At t = 2,

y'= 52 - 32(2) = 52 - 64 = - 12ft/s.

Instantaneous velocity at t=2 is given as -12ft/s.

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