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3 years ago

28:22:24 ratio simplyfied

Mathematics

2 answers:

Verdich [7]3 years ago

6 0

You just have to find the common multiple, which is two. Therefore, the ratio simplified is 14:11:12

worty [1.4K]3 years ago

4 0

28:22:24
divide by two as that is the common multiple for all of them
14:11:12

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Order the following from least to greatest. 3.1, 118%, 5/12, 0.06, 2/7

photoshop1234 [79]

Answer:

0.06, 2/7, 5/12, 118%, 3.1

Step-by-step explanation:

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3 years ago

The denarius was a unit of currency in ancient rome. Suppose it costs the roman government 101010 denarii per day to support 444

givi [52]

the correct question is

The denarius was a unit of currency in ancient rome. Suppose it costs the roman government 10 denarii per day to support 4 legionaries and 4 archers. It only costs 5 denarii per day to support 2 legionaries and 2 archers. Use a system of linear equations in two variables. Can we solve for a unique cost for each soldier?

Let

x-------> the cost to support a legionary per day

y-------> the cost to support an archer per day

we know that

4x+4y=10 ---------> equation 1

2x+2y=5 ---------> equation 2

If you multiply equation 1 by 2

2*(2x+2y)=2*5-----------> 4x+4y=10

equation 1 and equation 2 are the same

The system has infinite solutions-------> Is a consistent dependent system

therefore

the answer is

We cannot solve for a unique cost for each soldier, because there are infinite solutions.

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2 years ago

Explorers on a planet with a thin atmosphere want to measure the acceleration of gravity at its surface. So they use a spring gu

Murljashka [212]

Answer:

The value of $g_s$ is $g_s=\frac{30}{21} \:\frac{m}{s^2}$ .

Step-by-step explanation:

We know that the position function is given by

$s(t)=30t-\frac{g_s}{2} t^2$

Velocity is defined as the rate of change of position. Therefore,

$v(t)=\frac{d}{dt}(s(t))$

So, the velocity function is

$v(t)=\frac{d}{dt}(s(t))\\\\v(t)=\frac{d}{dt}(30t-\frac{g_s}{2} t^2)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(30t\right)-\frac{d}{dt}\left(\frac{g_s}{2}t^2\right)\\\\v(t)=30-g_st$

When a body reaches a vertical velocity of zero, this is the maximum height of the body and then gravity will take over and accelerate the object downward. Thus,

$v(t)=30-g_st=0\\\\g_s=\frac{30}{t}$

We know that the ball bearing reaches its maximum height 21 seconds after being launched (t = 21 s).

So, the value of $g_s$ is $g_s=\frac{30}{21} \:\frac{m}{s^2}$ .

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3 years ago

Cole drew a rectangle that is 3 meters by 10 meters. It has an area of 30 square meters and a perimeter of 26 meters. Which rect

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A rectangle with 8x5 dimensions

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2 years ago

Read 2 more answers

ONLY ANSWER NO WORK WILL MARK BRAINLIEST

Whitepunk [10]

Alright, here goes:
For this question, you are asked about the median and Interquartile Range(IQR), so first you need to understand what that means.
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2 years ago