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3 years ago

) Would you rather receive $1,500 a year for 12 years, or $1,200 a year for 20 years if the interest rate is 7%?

Mathematics

1 answer:

Umnica [9.8K]3 years ago

6 0

Answer:

$1200 should receive.

Step-by-step explanation:

To check whether to receive $1500 or $1200 just find the present values of each amount.

Present value of annuity $1500:

Present value of annuity = Annuity (P/A, 7%, 12)

Present value of annuity = 1500 (P/A, 7%, 12)

Present value of annuity = 1500 (7.9426)

Present value of annuity = $11913.9

Present value of annuity $1200:

Present value of annuity = Annuity (P/A, 7%, 20)

Present value of annuity = 1500 (P/A, 7%, 20)

Present value of annuity = 1500 (10.5940)

Present value of annuity = $15891

Thus, $1200 should receive.

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Amelia, Luis, Shauna, and Clarence used different approaches to solve the inequality

Mama L [17]

Answer:

Luis’s, because he flipped the inequality sign when he subtracted

Step-by-step explanation:

Given:

7.2b + 6.5 > 4.8b – 8.1.

Amelia started by subtracting 7.2b from both sides to get 6.5 > –2.4b – 8.1.

Amelia:

7.2b + 6.5 - 7.2b > 4.8b – 8.1 - 7.2b

6.5 > -2.4b - 8.1

Correct

Luis started by subtracting 4.8b from both sides to get 2.4b + 6.5 < – 8.1.

7.2b + 6.5 - 4.8b > 4.8b – 8.1 - 4.8b

2.4b + 6.5 > -8.1

Incorrect

Shauna started by subtracting 6.5 from both sides to get 7.2b > 4.8b – 14.6

7.2b + 6.5 - 6.5 > 4.8b – 8.1 - 6.5

7.2b > 4.8b - 14.6

Correct

Clarence started by adding 8.1 to both sides to get 7.2b + 14.6 > 4.8b.

7.2b + 6.5 + 8.1 > 4.8b – 8.1 + 8.1

7.2b + 14.6 > 4.8b

Correct

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2 years ago

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professor190 [17]

Answer:

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Step-by-step explanation:

You multiply 14 by n, which is 14n.

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ki77a [65]

Use the slope formula and slope-intercept form
y=mx+b
to find the equation.
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2 years ago

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Step-by-step explanation:

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Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

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3 years ago