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labwork [276]
3 years ago
5

________  _______ express the same relationship between two quantities.

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
7 0
<span>Equivalent ratios is the answer that fits this expresson. This means for example that 1 : 2 is the same as 2 : 4.</span>
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Margaux borrowed 63000 on jan. 16 2012 and paid the loan in full on may 13 2012 with 11% interest. Compute the simple interest u
sp2606 [1]

Answer:

  $2234.26

Step-by-step explanation:

The formula is ...

  i = Prt

Filling in the numbers, we have ...

  i = $63000·0.11·(118/366) ≈ $2234.26

The year beginning Jan 16, 2012 and ending on the same date in 2013 has 366 days, so we use that for the number of days in the year. Similarly, the day Feb 29, 2012 adds one day to the interest-earning period, so there are a total of 118 days between 1/16/2012 and 5/13/2012.

6 0
3 years ago
What is the y-coordinate of point D after a translation of<br> (x, y) - (x + 6,7 - 4)?
dybincka [34]

Hey There!!

Your answer will be 1.

Step-by-step explanation:

If the coordinates of the point D' are (x, y), then, after translation, it will be

D(x + 6, y - 4).

that is 6 units right and 4 units down

It is given that the point before translation is D'(3, 5).

So, x + 6 = 3 + 6 = 9 and

y - 4 = 5 - 4 = 1

So, the y coordinate of point D after translation is 1.

Hope This Helps!!!

By ♡Itsbrazts♡

8 0
3 years ago
Find the value of an investment of $15,000 for 13 years at an annual interest rate of 4.55% compounded continuously.
atroni [7]

Answer:

A = $27100.5

Step-by-step explanation:

We are given;

Principal: P = $15,000

Interest rate: r = 4.55% = 0.0455

Time in years: t = 13 years

The formula for the value of the investment after t years of continuous compounding is;

A = P(e^(rt))

A = 15000(e^(0.0455 × 13))

A = 15000 × 1.8067

A = $27100.5

4 0
3 years ago
What is the perimeter of a right triangle with a hypotenuse of length 13 and a leg of length 12?
Ratling [72]

12^2+b^2=13^2

144+b^2=169

subtract from both sides

b^2=25<--- (square root)

=5

I hope this helped :)

8 0
3 years ago
A mixture contains 40 ounces of glycol and water and it is 10% glycol the mixture is to be strengthened to 25% by adding glycol
Anuta_ua [19.1K]
Reasoning solution:

40 oz of 10% glycol contains 40*0.1=4 oz of glycol, and 40-4=36 oz. of water.

To make a 25% glycol using 36 oz of water will make a total volume of 36*(4/3)=48 oz.
Volume of pure glycol to be added = 48-40=8 oz

Therefore 8 oz of pure glycol must be added to the mixture to make a 25% solution.

Algebraic solution:
Let x = volume of pure glycol required in oz.
then 
0.10*40+1.00*x = 0.25*(40+x)
expand
4.0+x=10+0.25x
solve
(1-0.25x)=10-4=6
x=6/0.75=8 oz

Therefore 8 oz of pure glycol must be added to the mixture to make a 25% solution.
3 0
3 years ago
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