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4 years ago

PLEASE ANSWER! WILL GIVE BRAINLIEST!!!

Mathematics

2 answers:

bezimeni [28]4 years ago

8 0

Answer:

The starting balance of the 3rd year is $811.20.

The ending balance for the 3rd year is $843.65.

The interest earned in the 4th year is $33.75.

Step-by-step explanation:

I did it on edg.

riadik2000 [5.3K]4 years ago

4 0

Answer:

The starting balance of the 3rd year is $811.20.

The ending balance for the 3rd year is $843.65.

The interest earned in the 4th year is $33.75.

Step-by-step explanation:

750 × (1 + 4%)^n

750 × (1.04)^n

750 × 1.04² = 811.2

(End of second year, starting of third year)

750 × 1.04³ = 843.648

In third year: 843.648 - 811.2

= 32.448

750 × 1.04⁴ = 877.39392

During the fourth year:

877.39392 - 843.648

33.74592

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ElenaW [278]

<h2>Answer</h2>

$x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1$

Or as ordered pairs: $(\sqrt{6} ,0),(-\sqrt{6} ,0),(\sqrt{5} ,-1),(-\sqrt{5} ,-1)$

<h2>Explanation</h2>

Lets solve our system of equations step by step

$x^2+y^2=6$ equation (1)

$x^2-y=6$ equation (2)

1. Solve for $x^2$ in equation (2)

$x^2-y=6$

$x^2=6+y$ equation (3)

2. Replace equation (3) in equation (1) and solve for $y$

$x^2+y^2=6$

$6+y+y^2=6$

$y^2+y=0$

$y(y+1)=0$

$y=0$ or $y=-1$

3. Replace the values of $y$ in equation (3) and solve for $x$

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$x^2=6+y$

$x^2=6$

$x=\sqrt{6}$ or $x=-\sqrt{6}$

- For $y=-1$

$x^2=6+y$

$x^2=6-1$

$x^2=5$

$x=\sqrt{5}$ or $x=-\sqrt{5}$

So, the solutions of our system of equation are:

$x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1$

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