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horsena [70]
3 years ago
9

Haily paid $13 for 1 and 3/7 kf of sliced salami. What was the cost per kilogram of salami?

Mathematics
1 answer:
nikdorinn [45]3 years ago
4 0

13 divided by 1 and 3/7=9 and 1/10

$9.10

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Solve the equation 7j=11 <br> Enter each line of working as an equation.
NeTakaya

Answer:

j=1.571

Step-by-step explanation:

to get j by itself you have to divide 7 by 11

so,

11/7=j

j=1.571

7 0
3 years ago
The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s
Norma-Jean [14]

The positions when the particle reverses direction are:

s(t_1)=55ft\\\\s(t_2)=28ft

The acceleraton of the paticle when reverses direction is:

a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

s(t)=2t^{3}-21t^{2}+60t+3

So,

- Derivating to get the velocity function, we have:

v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42

Now, substituting the times to calculate the accelerations, we have:

a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}

Now, substitutitng the times to calculate the positions, we have:

s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft

Have a nice day!

3 0
3 years ago
What is the perimeter of ^PQR with vertices P(-2,9) Q(7,-3) and R(-2,-3) in the coordinate plane?
e-lub [12.9K]
Using the distance formula we can determine the perimeter of PQR.

Distance formula: d=√(x-x1)^2+(y-y1)^2
distance of PR: √(-2+2)^2+(9+3)^2= 12
distance of QR: √(7+2)^2+(-3+3)^2=9
distance of PQ: √(7+2)^2+(-3-9)^2= 15

Perimeter: 12+9+15=36
5 0
3 years ago
Still in class please need this fast
ivanzaharov [21]

Answer:

12 km

Step-by-step explanation:

C is 5 km

4+3+5=12

pls brainliest

7 0
2 years ago
Read 2 more answers
Which equation represents a circle that contains the point (-5. -3) and has a center at (-2, 1)?
nlexa [21]
The answer for this is b. Trust me on this
4 0
3 years ago
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