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3 years ago

What equation represents the slope intercept form of the line below Y intercept =(0,-2) slope = 2/3

Mathematics

2 answers:

Vikki [24]3 years ago

5 0

Answer: $y = \frac{2}{3} x - 2$

Step-by-step explanation:

The slope intercept form is y = mx + b

Where m is our slope and b is our y-intercept

We're given a slope of 2/3

We're given a y-intercept of -2

Substituting into y = mx + b gives

$y = \frac{2}{3} x - 2$

slamgirl [31]3 years ago

5 0

Answer:

y = (2/3)x - 2

Step-by-step explanation:

Use the slope-intercept form y = mx + b. Here b = -2 and m = 2/3. Making these replacements, we get:

y = (2/3)x - 2

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What is the equation of a line that passes through the points (–3, 4) and (2, 8)?

sveta [45]

Answer:

Step-by-step explanation:

4-9= -5

2-3 = -1

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3 0

3 years ago

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Explain when the congruent complements theorem would be appropriate to use in a proof.

Natasha2012 [34]

The congruent complements theorem 2 angles are complements of the same angle (or of congruent angles), then the two angles are congruent.

7 0

4 years ago

Find two consecutive odd integers whose sum is 196

WINSTONCH [101]

n, n + 2 - two consecutive odd integers

196 - the sum

The equation:

n + (n + 2) = 196

n + n + 2 = 196 |subtract 2 from both sides

2n = 194 |divide both sides by 2

n = 97

n + 2 = 97 + 2 = 99

Answer: 97, 99.

3 0

4 years ago

The height (feet) of an object moving vertically is given by s= (-16t^2)+(208t)+(156), where t is in seconds.

OverLord2011 [107]

Answer:

The object's velocity at t = 7 is -16 ft/s

The maximum height is 832 ft and it occurs when $t=\frac{13}{2}$

Step-by-step explanation:

From the information given, the equation of motion of the object is

$s(t)= -16t^2+208t+156$

For any equation of motion s(t), the instantaneous velocity at time t is given by

$v(t)=\frac{ds}{dt}$

(a) To find the object's velocity at t = 7, you must:

$v(t)=\frac{d}{dt}(-16t^2+208t+156)\\\\v(t)=-\frac{d}{dt}\left(16t^2\right)+\frac{d}{dt}\left(208t\right)+\frac{d}{dt}\left(156\right)\\\\v(t)=-32t+208+0\\\\v(t)=-32t+208$

Next, we evaluate when t = 7

$v(7)=-32(7)+208=-224+208\\\\v(7)=-16$

The object's velocity at t = 7 is -16 ft/s

(b) To find the maximum of a function we always use the derivative of the function and we set it equal zero to find the critical points.

To find the maximum height and when it occurs, we set the velocity function equal to 0 and solve for t.

$-32t+208=0\\\\-32t+208-208=0-208\\\\-32t=-208\\\\\frac{-32t}{-32}=\frac{-208}{-32}\\\\t=\frac{13}{2}$

Next, we substitute this value into the equation of motion to find the maximum height

$s(\frac{13}{2})= -16(\frac{13}{2})^2+208(\frac{13}{2})+156\\\\s(\frac{13}{2})=-676+1352+156)=832$

The maximum height is 832 ft and it occurs when $t=\frac{13}{2}$

6 0

3 years ago

PLEASE HELP WILL GIVE BRAINLIEST TO CORRECT ANSWER

ioda

Answer:

Step-by-step explanation:

h(x) = 7 - x - 2x^5

h(-1) = 7 + 1 - 2(-1)^5

h(-1) = 7 + 1 + 2

h(-1) = 10

Now differentiate h(x)

h'(x) = -1 - 2*5*x^4

h'(x) = -1 - 10x^4

h'(x) = -1 - 10(x)^4

h'(x) = -1 - 10 (x)^4

h'(-1) = -1 - 10(10000)

h'(-1) = -1 - 100000

h'(-1) = - 99999

3 0

3 years ago