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Aleksandr [31]
3 years ago
9

Item 8

Mathematics
1 answer:
ludmilkaskok [199]3 years ago
8 0
So, I didn't quite understand your question clearly, but I'll try my best to answer it.

So, the first question is quite simple: Did the runner have a % decrease in his time or an increase.
Well, 7:45 is minor than 5:51. So, his time decreased.

2nd question:
The percent of change. Well, first things first 7:45 has 2 different units, wich are minutes and seconds, I suppose. Same goes for 5:51.
So let's put everything in seconds. multiply both the 7 and 5 for 60(for every minute has 60 seconds.)

That means:
7*60= 420+45 = 465.
5*60=300+51=351.

Now, let's do the math itself:
465 -> 100%
351 -> x%
 Is equal to about  75,5%
How much did it decrease then?
(75,5-100)%= 24,5%, approximately

You might be interested in
Help please:))
Blababa [14]

Answer:

a) Because this asks about the radius and height, I assume that we are talking about a cylinder shape.

Remember that for a cylinder of radius R and height H the volume is:

V = pi*R^2*H

And the surface will be:

S = 2*pi*R*H + pi*R^2

where pi = 3.14

Here we know that the volume is 1000cm^3, then:

1000cm^3 = pi*R^2*H

We can rewrite this as:

(1000cm^3)/pi = R^2*H

Now we can isolate H to get:

H = (1000cm^3)/(pi*R^2)

Replacing that in the surface equation, we get:

S = 2*pi*R*H + pi*R^2

S = 2*pi*R*(1000cm^3)/(pi*R^2) + pi*R^2

S = 2*(1000cm^3)/R + pi*R^2

So we want to minimize this.

Then we need to find the zeros of S'

S' = dS/dR = -(2000cm^3)/R^2 + 2*pi*R = 0

So we want to find R such that:

2*pi*R = (2000cm^3)/R^2

2*pi*R^3 = 2000cm^3

R^3 = (2000cm^3/2*3.14)

R = ∛(2000cm^3/2*3.14) = 6.83 cm

The radius that minimizes the surface is R = 6.83 cm

With the equation:

H = (1000cm^3)/(pi*R^2)

We can find the height:

H = (1000cm^3)/(3.14*(6.83 cm)^2) =  6.83 cm

(so the height is equal to the radius)

b) The surface equation is:

S = 2*pi*R*H + pi*R^2

replacing the values of H and R we get:

S = 2*3.14*(6.83 cm)*(6.83 cm) + 3.14*(6.83 cm)^2 = 439.43 cm^2

c) Because if we pack cylinders, there is a lot of space between the cylinders, so when you store it, there will be a lot of space that is not used and that can't be used for other things.

Similarly for transport problems, for that dead space, you would need more trucks to transport your ice cream packages.

3 0
3 years ago
Find area and perimeter of the rectangle or square ​
Radda [10]

Answer:

Area:702

perimeter: 114

Step-by-step explanation:

24x27=648

6x9=54

648+54=702

27+24+18+6+9+30=114

4 0
2 years ago
Suppose an ant walks counterclockwise on the unit circle from the point to the endpoint of the radius that forms an angle of rad
serg [7]

Answer:

\frac{4\pi}{3}

Step-by-step explanation:

The original questions is suppose an ant walks counterclockwise on a unit circle from the point (1,0) to the endpoint of the radius that forms an angle of 240 degrees with the positive horizontal axis.

To find the distance ant walked we find the arc length of the sector with central angle 240 degree and radius =1 (unit circle)

arc length of a sector =\frac{centralangle}{360} \cdot 2\pi r

arc length of a sector =\frac{240}{360} \cdot 2\pi(1)

arc length of a sector =\frac{2}{3} \cdot 2\pi

\frac{4\pi}{3}

6 0
3 years ago
Twice a number added to a smaller number is 5. The difference of 5 times the smaller number and the larger number is 3. Let x re
Sergio [31]

Answer:

1) 2Y+X=5

2) 5X-Y=3

6 0
3 years ago
Find all solutions of the given system of equations (If the system is infinite many solution, express your answer in terms of x)
lisov135 [29]

Answer:

(a) The system of the equations \left \{ {2x-3y\:=3} \atop {4x-6y\:=3}} \right. has no solution.

(b) The system of the equations \left \{ {4x-6y\:=10} \atop {16x-24y\:=40}} \right. has many solutions y=\frac{2x}{3}-\frac{5}{3}

Step-by-step explanation:

(a) To find the solutions of the following system of equations \left \{ {2x-3y\:=3} \atop {4x-6y\:=3}} \right. you must:

Multiply 2x-3y=3 by 2:

\begin{bmatrix}4x-6y=6\\ 4x-6y=3\end{bmatrix}

Subtract the equations

4x-6y=3\\-\\4x-6y=6\\------\\0=-3

0 = -3 is false, therefore the system of the equations has no solution.

(b) To find the solutions of the system \left \{ {4x-6y\:=10} \atop {16x-24y\:=40}} \right. you must:

Isolate x for 4x-6y=10

x=\frac{5+3y}{2}

Substitute x=\frac{5+3y}{2} into the second equation

16\cdot \frac{5+3y}{2}-24y=40\\8\left(3y+5\right)-24y=40\\24y+40-24y=40\\40=40

The system has many solutions.

Isolate y for 4x-6y=10

y=\frac{2x}{3}-\frac{5}{3}

3 0
3 years ago
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