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Troyanec [42]
3 years ago
10

3200 dollars is placed in an account with an annual interest rate of 8.25%. To the nearest tenth of a year, how long will it tak

e for the account value to reach 9600 dollars?
Mathematics
1 answer:
frozen [14]3 years ago
6 0

It takes 24 years for the account value to reach the $9600.

<u>Step-by-step explanation:</u>

<u>From the given data, it can be determined that :</u>

  • Amount = $9600
  • Principal = $3200
  • Rate = 8.25% = 0.0825

Number of years should be calculated.

<u>To find the interest :</u>

Amount = Principal + Interest

Therefore, the interest can be calculated by the formula,

Interest = Amount - Principal

⇒ 9600 - 3200

⇒ 6400

The interest needed is 6400.

<u>To find the number of years required to reach this interest :</u>

Use the formula for simple interest,

Simple interest = Pnr /100

where,

  • P is the principal
  • n is the number of years
  • r is the rate

Substitute all the given values in the simple interest formula,

Simple interest = 3200 × n × 0.0825

6400 = 264 n

n = 6400 / 264

n = 24.2 which is approximately 24 years.

Therefore, it takes 24 years for the account value of 3200 dollars to reach 9600 dollars.

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Tank A capacity = 550 gallons       66% x 550 gallons = 363 gallons in Tank A

How about the water in tank B?

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2 years ago
Find the value of each variable.
Harrizon [31]

Step-by-step explanation:

b is per the identity of angles on parallel lines when intersected by one inclined line the same as the 40° angle.

so,

b = 40°

due to the parallel nature of the 2 lines there is a symmetry effect for such shapes inscribed a circle. the upper and the lower triangle must be similar. and when applying a vertical line through the central crossing point, everything to the left is mirrored by everything on the right.

so, angle c must be equal to angle b.

c = 40°

and as the sum of all angles in a triangle is always 180°, d is then

d = 180 - 40 - 40 = 100°

the interior angle of the arc angle a is the supplementary angle of d (together they are 180°), because together with d they cover the full down side of the top-left to bottom-right line.

interior angle to a = 180 - 100 = 80°

due to the symmetry again, the arc angle opposite to a is the same as a.

as we know, the interior angle to a pair of opposing arc angles is the mean value of the 2 angles.

so, we have

(a + a)/2 = 80

2a/2 = 80

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there might (and actually should) be some more direct approaches for "a" out of the other pieces of information, but that was the most straight one right out of my mind, and I don't spend time on finding additional shortcuts, when I have already a working approach.

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CALCULUS - Find the values of in the interval (0,2pi) where the tangent line to the graph of y = sinxcosx is
Rufina [12.5K]

Answer:

\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}

Step-by-step explanation:

We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.

Since the tangent line is horizontal, this means that our derivative at those points are 0.

So, first, let's find the derivative of our function.

y=\sin(x)\cos(x)

Take the derivative of both sides with respect to x:

\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]

We need to use the product rule:

(uv)'=u'v+uv'

So, differentiate:

y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]

Evaluate:

y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))

Simplify:

y'=\cos^2(x)-\sin^2(x)

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

0=\cos^2(x)-\sin^2(x)

Now, let's solve for x. First, we can use the difference of two squares to obtain:

0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))

Zero Product Property:

0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)

Solve for each case.

Case 1:

0=\cos(x)-\sin(x)

Add sin(x) to both sides:

\cos(x)=\sin(x)

To solve this, we can use the unit circle.

Recall at what points cosine equals sine.

This only happens twice: at π/4 (45°) and at 5π/4 (225°).

At both of these points, both cosine and sine equals √2/2 and -√2/2.

And between the intervals 0 and 2π, these are the only two times that happens.

Case II:

We have:

0=\cos(x)+\sin(x)

Subtract sine from both sides:

\cos(x)=-\sin(x)

Again, we can use the unit circle. Recall when cosine is the opposite of sine.

Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.

At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).

At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.

Therefore, our solution set is:

\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}

And we're done!

Edit: Small Mistake :)

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