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Svetlanka [38]
3 years ago
7

What is the slope of the line with equation y−3=(x−2)?

Mathematics
2 answers:
TEA [102]3 years ago
4 0

Answer:

1

Step-by-step explanation:

y - 3 = (x - 2)

Solve for y. Then when the equation is in the slope-intercept form, y = mx + b, m is the slope.

y - 3 = x - 2

Add 3 to both sides.

y = x + 1

Compare to y = mx + b: y = 1x + 1

m = 1

Answer: The slope is 1.

grin007 [14]3 years ago
3 0

Answer:

1

Step-by-step explanation:

The slope of the line (when in point-slope form) will be the coefficient of the term on the right side of the equation. In this case that is 1.

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A school administrator believes that the mean class gpas for a given course are higher than the preferred mean of 2.75 at a sign
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The mean GPA is greater than 2.75.

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2 years ago
Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.
finlep [7]

Answer:

a) v = \frac{[L]}{[T]} = LT^{-1}

b) a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

c) \int v dt = s(t) = [L]=L

d) \int a dt = v(t) = [L][T]^{-1}=LT^{-1}

e) \frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

v = \frac{ds}{dt}

a= \frac{dv}{dt}

Part a

If we do the dimensional analysis for v we got:

v = \frac{[L]}{[T]} = LT^{-1}

Part b

For the acceleration we can use the result obtained from part a and we got:

a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

Part c

From definition if we do the integral of the velocity respect to t we got the position:

\int v dt = s(t)

And the dimensional analysis for the position is:

\int v dt = s(t) = [L]=L

Part d

The integral for the acceleration respect to the time is the velocity:

\int a dt = v(t)

And the dimensional analysis for the position is:

\int a dt = v(t) = [L][T]^{-1}=LT^{-1}

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

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3 years ago
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