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Irina18 [472]
3 years ago
11

How do I andser this. I’m stuck

Mathematics
1 answer:
Mars2501 [29]3 years ago
8 0
A= 3:5
b= 12:20
c= yes because 3” 3+2=5 n in the beginning of the problem it said end of season
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Each container must hold exactly 1 litre of water Each container must have a minimum surface area
Arada [10]

The containers must be spheres of radius = 6.2cm

<h3>How to minimize the surface area for the containers?</h3>

We know that the shape that minimizes the area for a fixed volume is the sphere.

Here, we want to get spheres of a volume of 1 liter. Where:

1 L = 1000 cm³

And remember that the volume of a sphere of radius R is:

V = \frac{4}{3}*3.14*R^3

Then we must solve:

V = \frac{4}{3}*3.14*R^3 = 1000cm^3\\\\R =\sqrt[3]{  (1000cm^3*\frac{3}{4*3.14} )} = 6.2cm

The containers must be spheres of radius = 6.2cm

If you want to learn more about volume:

brainly.com/question/1972490

#SPJ1

6 0
2 years ago
In the isosceles trapezoid below,<br> X =<br> 5x+ 15°<br> 7x-11°
tangare [24]

Answer:

x = 13°

Step-by-step explanation:

5x+15°=7x-11°

5x-7x = -11°-15°

-2x = -26° divide both sides by -2

x = 13°

5x+15° = 5×13°+15° = 80°

7x-11° = 7×13° -11°. = 80°

6 0
2 years ago
A spring with a spring constant k of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. It is then st
shutvik [7]

Answer:

Equation of movement y(t)

y(t)=sin(10t-\pi/2)

Amplitude: 1 inch

Period: 0.628 seconds

Step-by-step explanation:

If there is no friction, the amplitude will be the length it was streched from the equilibrium. In this case, this value is 1 inch

A=1\,inch

The period depends on the mass and the spring constant.

The formula for the period is:

T=\frac{1}{f} =\frac{2/pi}{\omega} =2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{1}{100}}=2\pi*0.1=0.628\, s

The model y(t) for the movement of the mass-spring system is

y(t)=A\cdot sin(\omega t +\phi)\\\\A=1\\\\\omega=\sqrt{k/m} =\sqrt{100/1}=10\\\\y(0)=Asin(\phi)=-A\\\\ \phi=arcsin(-1)=-1.571=-\pi/2\\\\\\y(t)=sin(10t-\pi/2)

8 0
3 years ago
Give the following equations determine if the lines are parallel perpendicular or neither
GaryK [48]

In order to determine whether the equations are parallel, perpendicular, or neither, let's simply each equation into a slope-intercept form or basically, solve for y.

Let's start with the first equation.

\frac{6x-5y}{2}=x+1

Cross multiply both sides of the equation.

6x-5y=2(x+1)6x-5y=2x+2

Subtract 6x on both sides of the equation.

6x-5y-6x=2x+2-6x-5y=-4x+2

Divide both sides of the equation by -5.

-\frac{5y}{-5}=\frac{-4x}{-5}+\frac{2}{-5}y=\frac{4}{5}x-\frac{2}{5}

Therefore, the slope of the first equation is 4/5.

Let's now simplify the second equation.

-4y-x=4x+5

Add x on both sides of the equation.

-4y-x+x=4x+5+x-4y=5x+5

Divide both sides of the equation by -4.

\frac{-4y}{-4}=\frac{5x}{-4}+\frac{5}{-4}y=-\frac{5}{4}x-\frac{5}{4}

Therefore, the slope of the second equation is -5/4.

Since the slope of each equation is the negative reciprocal of each other, then the graph of the two equations is perpendicular to each other.

5 0
1 year ago
1a) What percentage of Priya’s distance did Tyler walk?
umka2103 [35]
I’m not sure I only answer these for the points :)
7 0
3 years ago
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