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IgorLugansk [536]
3 years ago
10

What is the product of 5/12 x 3

Mathematics
1 answer:
Sliva [168]3 years ago
4 0
5/12 x 3/1 = 15/12 = 1 3/12
Simplified
1 1/4
Not simplified
1 3/12
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Using the graphing function on your calculator, what is the solution to the system of equations shown below?
SpyIntel [72]
Answer:
B, x=3 y=5

Step-by-step explanation:
x=-7+2y
3y+2x=21
Substitute the x into the other equation
3y+2(-7+2y)=21
y=5
Now put the value into a desired equation
x=-7+2(5)
x=3.
The answer is x=3 and y=5

I hope this helped!
5 0
3 years ago
Which equation describes the same line as y – 6 = –4(x + 1)?
Lorico [155]
I would say c but that is just a guess

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Find the precent of change of 15 yards to 18 yards
Delvig [45]
Divide the current with the previous
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The burj khalifa a building in dubai united arab emirates is about 828 meters tall
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The Burj Khalifa, a building in Dubai, United Arab Emirates, is about 828 meters tall. Brody is making a scale model of this building, using the scale 500 meters : 1 meter.
6 0
3 years ago
An equation for the depreciation of a car is given by y=A(1-r)t where y=current value of the car.A=original cost r=rate of depre
Ann [662]
<h3>Answer: Approximately 6.58 years old</h3>

The more accurate value is 6.57881347896059, which you can round however you need. I picked two decimal places.

==================================================

Explanation:

Let's pick a starting value of the car. It doesn't matter what the starting value, but it might help make the problem easier. Let's say A = 1000. Half of that is 1000/2 = 500.

So we want to find out how long it takes for the car's value to go from $1000 to $500 if it depreciates 10% per year.

The value of r is r = 0.10 as its the decimal form of 10%

t is the unknown number of years we want to solve for

---------------------------

y = A(1 - r)^t

500 = 1000(1 - 0.1)^t

500 = 1000(0.9)^t

1000(0.9)^t = 500

0.9^t = 500/1000

0.9^t = 0.5

log( 0.9^t ) = log( 0.5 )

t*log( 0.9 ) = log( 0.5 )

t = log( 0.5 )/log( 0.9 )

t = 6.57881347896059

Note the use of logs to help us isolate the exponent.

6 0
3 years ago
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