Bobby has 3/4 of a bag of chips.He

Is 28.256 a positive or negative notation?

WARRIOR [948]

Answer:

To change a number from scientific notation to standard form, move the decimal point to the left (if the exponent of ten is a negative number), or to the right (if the exponent is positive). You should move the point as many times as the exponent indicates. Do not write the power of ten anymore.

HOPE THIS HELPED!!!!!!!!!!!!!XDDDDDD

7 0

3 years ago

What is the equation of a line that has a slope of 2, and contains the point (4,6)?

Tems11 [23]

Answer:

y= 2x -2

Step-by-step explanation:

You would use point- slope formula. y-y = m(x-x). With your numbers it would end up looking like this ----> y-6 = 2(x-4), then do the distributive property which then gives you ----> y-6 = 2x - 8, then subtract 6 on both sides and you get ---> y = 2x - 2. Hope this helped!!!!

3 0

3 years ago

Bethany is baking cookies for her school's bake sale. She plans to bake more than 185 cookies, in batchs or 12 cookies. so far,

Aleks [24]

Answer:

48+12b≥185

Step-by-step explanation:

Inequality: 48+12b≥185

Why? she has already backed 48 cookies so you just need to find how many more she needs to bake. (she doesn't need 185 cookies anymore since she's already baked 48)

You don' t need to solve the question,

But here is how you do...

12b≥137

b≥11.42 (roughly)

5 0

4 years ago

Find the area of this semi-circle with radius, r = 66cm. Give your answer rounded to 1 DP.

Darya [45]

Answer:

414.857143

Step-by-step explanation:

r=66

=2*66*22/7

=414.857143

4 0

2 years ago

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:

$0< t$

For this case if we select two values let's say 0.25 and 0.5 we see that

$v(0.25) =6.1875, v(0.5)=3.75$

And we see that for $a=0.5 >0.25=b$ we have that f(b)>f(a) so then the function is decreasing on this case.

$1$

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

$v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2$

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

$2$

We see that the function would be increasing.

$3$

For this interval we will see that for any two points a,b with $a>b$ we have $f(a)>f(b)$ for example let's say a=3 and b =4

$f(a=3) =0 , f(b=4) =9 , f(b)>f(a)$

The particle is moving to the right then the velocity is positive so then the answer for this case is: $0$

Part 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:

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

5 0

3 years ago