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1 year ago

Jason’s savings account has a balance of $2179. After 5 years , what will the amount of interest be at 6% compounded quarterly?

Mathematics

1 answer:

satela [25.4K]1 year ago

5 0

Answer:

$755.80

Step-by-step explanation:

Determine the compound amount first and then subtract the principal from it, to find the amount of interest.

The compound amount formula is A = P (1 + r/n)^(nt), where

P is the initial principal, r is the interest rate as a decimal fraction, n is the number of compounding periods per year, and t is the number of years. Here, P = $2179; t = 5 yrs; r = 0.06; and n = 4 (quarterly compounding).

We get:

A = $2179(1 + 0.06/4)^(4*5), or $2179(1.015)^20, or $2179(1.347) = $2937.80.

The compound amount is $2934.80. Subtracting the $2179 principal results in the interest earned: $755.80.

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1 year ago

The general form of the equation of a circle is x2 + y2 + 42x + 38y − 47 = 0. The equation of this circle in standard form is (x

almond37 [142]

Step 1

we know that

The equation of a circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center of the circle

r is the radius of the circle

In this problem we have

$x^{2} +y^{2} +42x+38y-47=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+42x)+(y^{2}+38y)=47$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+42x+21^{2})+(y^{2}+38y+19^{2})=47+21^{2}+19^{2}$

$(x^{2}+42x+21^{2})+(y^{2}+38y+19^{2})=849$

Rewrite as perfect squares

$(x+21)^{2}+(y+19)^{2}=849$

The center of the circle is the point $(-21,-19)$

The radius of the circle is $\sqrt{849}\ units$

The answer Part a) is

The equation of the circle in standard form is equal to

$(x+21)^{2}+(y+19)^{2}=849$

The answer Part b) is

The center of the circle is the point $(-21,-19)$

The answer Part c) is

The radius of the circle is $\sqrt{849}\ units$

Let's verify each case to determine the solution of the second part of the problem

Step 2

we have

$x^{2} +y^{2} +60x+14y+98=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+60x)+(y^{2}+14y)=-98$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+60x+30^{2})+(y^{2}+14y+7^{2})=-98+30^{2}+7^{2}$

$(x^{2}+60x+30^{2})+(y^{2}+14y+7^{2})=851$

Rewrite as perfect squares

$(x+30)^{2}+(y+7)^{2}=851$

The radius of the circle is $\sqrt{851}\ units$

therefore

This circle does not have the same radius of the circle above

Step 3

we have

$x^{2} +y^{2} +44x-44y+117=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+44x)+(y^{2}-44y)=-117$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+44x+22^{2})+(y^{2}-44y+22^{2})=-117+22^{2}+22^{2}$

$(x^{2}+44x+22^{2})+(y^{2}-44y+22^{2})=851$

Rewrite as perfect squares

$(x+22)^{2}+(y-22)^{2}=851$

The radius of the circle is $\sqrt{851}\ units$

therefore

This circle does not have the same radius of the circle above

Step 4

we have

$x^{2} +y^{2} -38x+42y+74=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}-38x)+(y^{2}+42y)=-74$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}-38x+19^{2})+(y^{2}+42y+21^{2})=-74+19^{2}+21^{2}$

$(x^{2}-38x+19^{2})+(y^{2}+42y+21^{2})=728$

Rewrite as perfect squares

$(x-19)^{2}+(y+21)^{2}=728$

The radius of the circle is $\sqrt{728}\ units$

therefore

This circle does not have the same radius of the circle above

Step 5

we have

$x^{2} +y^{2} -50x-30y+1=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}-50x)+(y^{2}-30y)=-1$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}-50x+25^{2})+(y^{2}-30y+15^{2})=-1+25^{2}+15^{2}$

$(x^{2}-50x+25^{2})+(y^{2}-30y+15^{2})=849$

Rewrite as perfect squares

$(x-25)^{2}+(y-15)^{2}=849$

The radius of the circle is $\sqrt{849}\ units$

therefore

This circle has the same radius of the circle above

therefore

The answer is

$x^{2} +y^{2} -50x-30y+1=0$ -----> has the same radio that the circle above

7 0

1 year ago

Fernando’s birthday is on March 21st. On March 9th, Fernando’s father said that he will give Fernando $5 on each date that is di

valentina_108 [34]

Answer:

15 years

Step-by-step explanation:

Days divisible by 4 between 9th and 21st (12, 16,20). = 3

Assume, New age = x

(3 * 5) + (10 + x) = 40

15 + 10 + x = 40

25 + x = 40

x = 40 - 25

x = 15 years

Hence, Fernando will be 15 years

7 0

1 year ago