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1 year ago

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An account is opened with $7,595.96 with a rate of increase of 2% per year. After 1 year, the bank account contains $7,746.90. A

ssuming no deposits or withdrawals are made, which equation can be used to find y, the amount of money in the account after x years

1 answer:

wolverine [178]1 year ago

3 0

Answer:

y = 7,595.96(1.02)x

Step-by-step explanation:

y = 7,595.96(1.02)x

a normal linear equation is basically y = x + b

in this case,

y = amount of money in the bank account
x = number of years
b is not required because there are no new deposits or withdrawals from the account

Since the accounts earns 2% interest per year, then you need to multiply the original amount by 1.02. As more years pass, the account will increase by 1.02, so the account increases by 1.02x. Since the original amount is $7,595.96, that will be our starting point.

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300-7[4(3+5)]+3 to the 3rd power

Mila [183]

Answer:

The answer is 103

Step-by-step explanation:

The given expression is:

300-7[4(3+5)]+3 to the 3rd power

3 to the 3rd power means 3^3

Therefore,

300-7[4(3+5)]+3^3

First we will solve the round bracket and find the cube

300-7[4(8)]+27

Now we will solve square bracket

300-7[32]+ 27

300-224+27

76+27

103

Thus the answer we get is 103....

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2 years ago

Learning Task 3. Find the equation of the line. Do it in your notebook.

Wewaii [24]

Answer:

1) The equation of the line in slope-intercept form is $y = 5\cdot x +9$ . The equation of the line in standard form is $-5\cdot x + y = 9$ .

2) The equation of the line in slope-intercept form is $y = \frac{2}{5}\cdot x +\frac{14}{5}$ . The equation of the line in standard form is $-2\cdot x +5\cdot y = 14$ .

3) The equation of the line in slope-intercept form is $y = 3\cdot x +4$ . The equation of the line in standard form is $-3\cdot x +y = 4$ .

4) The equation of the line in slope-intercept form is $y = 2\cdot x + 6$ . The equation of the line in standard form is $-2\cdot x +y = 6$ .

5) The equation of the line in slope-intercept form is $y = \frac{5}{6}\cdot x -\frac{7}{6}$ . The equation of the line in standard from is $-5\cdot x + 6\cdot y = -7$ .

Step-by-step explanation:

1) We begin with the slope-intercept form and substitute all known values and calculate the y-intercept: ( $m = 5$ , $x = -1$ , $y = 4$ )

$4 = (5)\cdot (-1)+b$

$4 = -5 +b$

$b = 9$

The equation of the line in slope-intercept form is $y = 5\cdot x +9$ .

Then, we obtain the standard form by algebraic handling:

$-5\cdot x + y = 9$

The equation of the line in standard form is $-5\cdot x + y = 9$ .

2) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = 3$ , $y_{1} = 4$ , $x_{2} = -2$ , $y_{2} = 2$ )

$3\cdot m + b = 4$ (Eq. 1)

$-2\cdot m + b = 2$ (Eq. 2)

From (Eq. 1), we find that:

$b = 4-3\cdot m$

And by substituting on (Eq. 2), we conclude that slope of the equation of the line is:

$-2\cdot m +4-3\cdot m = 2$

$-5\cdot m = -2$

$m = \frac{2}{5}$

And from (Eq. 1) we find that the y-Intercept is:

$b=4-3\cdot \left(\frac{2}{5} \right)$

$b = 4-\frac{6}{5}$

$b = \frac{14}{5}$

The equation of the line in slope-intercept form is $y = \frac{2}{5}\cdot x +\frac{14}{5}$ .

Then, we obtain the standard form by algebraic handling:

$-\frac{2}{5}\cdot x +y = \frac{14}{5}$

$-2\cdot x +5\cdot y = 14$

The equation of the line in standard form is $-2\cdot x +5\cdot y = 14$ .

3) By using the slope-intercept form, we obtain the equation of the line by direct substitution: ( $m = 3$ , $b = 4$ )

$y = 3\cdot x +4$

The equation of the line in slope-intercept form is $y = 3\cdot x +4$ .

Then, we obtain the standard form by algebraic handling:

$-3\cdot x +y = 4$

The equation of the line in standard form is $-3\cdot x +y = 4$ .

4) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = -3$ , $y_{1} = 0$ , $x_{2} = 0$ , $y_{2} = 6$ )

$-3\cdot m + b = 0$ (Eq. 3)

$b = 6$ (Eq. 4)

By applying (Eq. 4) on (Eq. 3), we find that the slope of the equation of the line is:

$-3\cdot m+6 = 0$

$3\cdot m = 6$

$m = 2$

The equation of the line in slope-intercept form is $y = 2\cdot x + 6$ .

Then, we obtain the standard form by algebraic handling:

$-2\cdot x +y = 6$

The equation of the line in standard form is $-2\cdot x +y = 6$ .

5) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = -1$ , $y_{1} = -2$ , $x_{2} = 5$ , $y_{2} = 3$ )

$-m+b = -2$ (Eq. 5)

$5\cdot m +b = 3$ (Eq. 6)

From (Eq. 5), we find that:

$b = -2+m$

And by substituting on (Eq. 6), we conclude that slope of the equation of the line is:

$5\cdot m -2+m = 3$

$6\cdot m = 5$

$m = \frac{5}{6}$

And from (Eq. 5) we find that the y-Intercept is:

$b = -2+\frac{5}{6}$

$b = -\frac{7}{6}$

The equation of the line in slope-intercept form is $y = \frac{5}{6}\cdot x -\frac{7}{6}$ .

Then, we obtain the standard form by algebraic handling:

$-\frac{5}{6}\cdot x +y =-\frac{7}{6}$

$-5\cdot x + 6\cdot y = -7$

The equation of the line in standard from is $-5\cdot x + 6\cdot y = -7$ .

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