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katrin2010 [14]
2 years ago
7

You invest $5000 into an account that has a 2.7% annual interest rate and is compounded quarterly. Approximately how long will i

t take until you have $7,500 in your account?​
Mathematics
1 answer:
allsm [11]2 years ago
4 0

Since the interest is compounded, we will have to use the compound interest formula.

We Weill plug 7500 in for A, because that's the amount of money that we want to have at the end of some amount of time.

5000 will go in for P because that's the starting amount.

2.7% will be converted into a decimal percentage form. You can do this by dividing by 100, which you will get .027, and then plug that in for r, the rate.

Since the interest is compounded quarterly, n = 4.

After a bit of number crunching, you will get to the point where you have to solve for an exponent. You can easily do this by using the natural log ln(). One property of logarithm is that you can take the exponent and place it in front of the log. Now you can divide both sides to separate and solve for t.

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Using the law of cosines, write an algebraic proof to show that the angles in an equilateral triangle must equal 60°. Use "∧" to
bekas [8.4K]

The algebraic proof shows that the angles in an equilateral triangle must equal 60° each

<h3>Laws of cosines </h3>

From the question, we are to use the law of cosines to write an algebraic proof that shows that the angles in an equilateral triangle must equal 60°.

Given any triangle ABC, the measures of angles A, B, and C by the law of cosines are

cos A = (b^2 + c^2 - a^2)/2bc

cos B= (a^2 + c^2 - b^2)/2ac

cos C = (a^2 + b^2 - c^2)/2ab  

Now, given that the triangle is equilateral, with each of the side lengths equal to s

That is, a = b = c = s

Then, we can write that

cos A = (s^2 + s^2 - s^2)/(2s×s)

cos A = (s^2 )/(2s^2)

cos A = 1/2

cos A = 0.5

∴ A = cos⁻¹(0.5)

A = 60°

Also

cos B = (s^2 + s^2 - s^2)/(2s×s)

cos B = (s^2 )/(2s^2)

cos B = 1/2

cos B = 0.5

∴ B = cos⁻¹(0.5)

B = 60°

and

cos C = (s^2 + s^2 - s^2)/(2s×s)

cos C = (s^2 )/(2s^2)

cos C = 1/2

cos C = 0.5

∴ C = cos⁻¹(0.5)

C = 60°

Thus,

A = 60°, B = 60° and C = 60°

Hence, the algebraic proof above shows that the angles in an equilateral triangle must equal 60° each.

Learn more on The law of cosines here: brainly.com/question/2866347

#SPJ1

3 0
2 years ago
-1/2 (10+1/4)
andriy [413]

Answer:

-5 1/8

Step-by-step explanation:

-5 1/8

7 0
2 years ago
What is the GCF of the expression 7xyz - 21xyz + 49yz + 14yz2?
nexus9112 [7]

Answer:

7yz

Step-by-step explanation:

You can take 7yz common from all the terms in the given expression

Answered by GAUTHMATH

8 0
2 years ago
Find the dimensions of the rectangle with largest area that can be inscribed in an equilateral triangle with sides of 1 unit, if
prohojiy [21]
<span>Maximum area = sqrt(3)/8 Let's first express the width of the triangle as a function of it's height. If you draw an equilateral triangle, then a rectangle using one of the triangles edges as the base, you'll see that there's 4 regions created. They are the rectangle, a smaller equilateral triangle above the rectangle, and 2 right triangles with one leg being the height of the rectangle and the other 2 angles being 30 and 60 degrees. Let's call the short leg of that triangle b. And that makes the width of the rectangle equal to 1 minus twice b. So we have w = 1 - 2b b = h/sqrt(3) So w = 1 - 2*h/sqrt(3) The area of the rectangle is A = hw A = h(1 - 2*h/sqrt(3)) A = h*1 - h*2*h/sqrt(3) A = h - 2h^2/sqrt(3) We now have a quadratic equation where A = -2/sqrt(3), b = 1, and c=0. We can solve the problem by using a bit of calculus and calculating the first derivative, then solving for 0. But since this is a simple quadratic, we could also take advantage that a parabola is symmetrical and that the maximum value will be the midpoint between it's roots. So let's use the quadratic formula and solve it that way. The 2 roots are 0, and 1.5/sqrt(3). The midpoint is (0 + 1.5/sqrt(3))/2 = 1.5/sqrt(3) / 2 = 0.75/sqrt(3) So the desired height is 0.75/sqrt(3). Now let's calculate the width: w = 1 - 2*h/sqrt(3) w = 1 - 2* 0.75/sqrt(3) /sqrt(3) w = 1 - 2* 0.75/3 w = 1 - 1.5/3 w = 1 - 0.5 w = 0.5 The area is A = hw A = 0.75/sqrt(3) * 0.5 A = 0.375/sqrt(3) Now as I said earlier, we could use the first derivative. Let's do that as well and see what happens. A = h - 2h^2/sqrt(3) A' = 1h^0 - 4h/sqrt(3) A' = 1 - 4h/sqrt(3) Now solve for 0. A' = 1 - 4h/sqrt(3) 0 = 1 - 4h/sqrt(3) 4h/sqrt(3) = 1 4h = sqrt(3) h = sqrt(3)/4 w = 1 - 2*(sqrt(3)/4)/sqrt(3) w = 1 - 2/4 w = 1 -1/2 w = 1/2 A = wh A = 1/2 * sqrt(3)/4 A = sqrt(3)/8 And the other method got us 0.375/sqrt(3). Are they the same? Let's see. 0.375/sqrt(3) Multiply top and bottom by sqrt(3) 0.375*sqrt(3)/3 Multiply top and bottom by 8 3*sqrt(3)/24 Divide top and bottom by 3 sqrt(3)/8 Yep, they're the same. And since sqrt(3)/8 looks so much nicer than 0.375/sqrt(3), let's use that as the answer.</span>
7 0
3 years ago
Read 2 more answers
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