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3 years ago

A = bh; A =96, b = 12 What does h equal?

1 answer:

6 0

The answer is 8. Wolfram|Alpha is great for these kinds of questions. Go to http://www.wolframalpha.com/ and math-y classes are much easier tbh

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You paint 1 2 12 wall in 1 4 14 hour. At that rate, how long will it take you to paint one wall

Rudiy27

Ok. well the answer 1/2 the hour you paint 1/2 in 1/4 so add 1/2+1/2=1 so it is 2 so 1/4+1/4

3 0

3 years ago

Read 2 more answers

Which equation represents a line that passes through (-2, 4) and has a slope of 1/2?

jok3333 [9.3K]

Answer:

$\large\boxed{y-4=\dfrac{1}{2}(x+2)\text{- point-slope form}}\\\boxed{y=\dfrac{1}{2}x+5\text{- slope-intercept form}}$

Step-by-step explanation:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)$

m - slope

We have the slope $m=\dfrac{1}{2}$ and the point $(-2, 4)$ .

Substitute:

$y-4=\dfrac{1}{2}(x-(-2))$

$y-4=\dfrac{1}{2}(x+2)$ - point-slope form

Convert to the slope-intercept form (y = mx + b):

$y-4=\dfrac{1}{2}(x+2)$ use the distributive property

$y-4=\dfrac{1}{2}x+1$ add 4 to both sides

$y=\dfrac{1}{2}x+5$ - slope-intercept form

3 0

3 years ago

List these numbers to least go greatest 0 , 8.5 , -3.4 , -1.5 , 8.45 ,1/2

elixir [45]

-3.4, -1.5, 0, 1/2, 8.5, and 8.45
Hope this helped:))

5 0

4 years ago

Read 2 more answers

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

vodka [1.7K]

Answer:

a) $v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

$x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10$

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals: