Broccoil cost 3.14$ a pound at the market if mary bought 2.5 pounds what was the cost

Consider the inequality x + 7 < 9. Select all the true statements.

alex41 [277]

Answer:

Step-by-step explanation:

here you go mate

step 1

x + 7 < 9 equation

step 2

x + 7 < 9 subtract -7

x+7(-7)<9(-7)

answer

x<2

3 0

3 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

15 POINTS-

Doss [256]

I think it’s $3.55
I think this because if u subtract 2.45- 45% = 1.10 sooo u will have to add $1.10 to 2.25 and that will add up the 45% that Walmart is paying with the original price.

8 0

3 years ago

Whoever checks my work and tell what I did wrong and the right answer will

gulaghasi [49]

Answer:

See below

Step-by-step explanation:

I believe that you only had to do letters F, H, and J. In that case, let's go over each one!

F: For isolating $d_{1}$ , we need to get rid of the 1/2 first. Let's multiply each side by 2:

$2*m = 2*\frac{1}{2}(d_{1}+ d_{2}) \\2m = d_{1}+ d_{2}$

After this, we just subtract $d_{2}$ from each side to get $d_{1}=2m- d_{2}$ . Dark Blue is correct! Let's now plug in those numbers below:

$d_{1}=2(10)- 13 = 20-13=7$

G: Let's isolate the vw^2 on one side by subtracting y from each side:

$vw^{2} +y-y=x-y\\vw^{2}=x-y$

Let's now divide each side by v, then put each side under a square root to get our final answer:

$\frac{vw^2}{v} = \frac{x-y}{v}\\ w^2 = \frac{x-y}{v}\\\sqrt{w^2} = \sqrt{\frac{x-y}{v}} \\w=\sqrt{\frac{x-y}{v}}$

Orange is correct! Again, let's solve the problem underneath:

w= $w=\sqrt{\frac{38-(-7)}{5}} = \sqrt{\frac{38+7}{5}} = \sqrt{\frac{45}{5}}=\sqrt{9}=3$

H: This one has some stuff that we haven't worked with quite yet (like terms), but our approach is the same: isolate c on one side of the equation.

$2a-2a+3c=17a-2a+21\\3c = 15a+21\\\frac{3c}{3} = \frac{15a+21}{3}\\c = \frac{15a}{3}+ \frac{21}{3} \\c = 5a+7$

Purple is correct! Let's solve the problem:

$c = 5(\frac{16}{5})+7 = 16+7 = 23$

7 0

3 years ago

Find the value of the lesser root of x2 + 12x + 35 = 0.

matrenka [14]

In quadratic equations,
a + b = -B
ab = C
where a and b are the roots, B is the second term, and C is the constant.
Substituting,
a + b = -12
ab = 35
The values of a and b from the equation is -7 and -5. Thus, the lesser root is -7.

8 0

3 years ago