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3 years ago

7

Justin deposited $2,000 into an account 5 years ago. Simple interest was paid on the account. He has just withdrawn $2,876. What

interest rate did he earn on the account?

1 answer:

sdas [7]3 years ago

6 0

Answer: $8.76\%$

Step-by-step explanation:

The formula to find the final amount after getting simple interest :

$A=P(1+rt)$ , where P is the principal amount , r is rate of interest ( in decimal )and t is time(years).

Given : Justin deposited $2,000 into an account 5 years ago.

i.e. P = $2,000 and t= 5 years

He has just withdrawn $2,876.

i.e. we assume that A = $2876

Now, Put all the values in the formula , we get

$(2876)=(2000)(1+r(5))\\\\\Rightarrow\ 1+5r=\dfrac{2876}{2000}\\\\\Rightarrow\ 1+5r=1.438\\\\\Righhtarrow\ 5r=0.438\\\\\Rightarrow\ r=\dfrac{0.438}{5}=0.0876$

In percent, $r=0.0876\times100=8.76\%$

hence, He earned $8.76\%$ of interest on account.

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The owner of the Outdoor Furniture Center decides to use the 120% markup. At the end on

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Answer: The total price of a bench is 1.008x.

Step-by-step explanation:

Since we have given that

Let the cost price be 'x'.

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So, Mark up value would be

$\dfrac{120}{100}x\\\\=1.20x$

Discount % = 20%

Amount of discount is given by

$\dfrac{20}{100}\times 1.2x\\\\=0.2\times 1.2x\\\\=0.24x$

So, it becomes,

Amount after discount is given by

$1.2x-0.24x\\\\=0.96x$

Sales tax = 5%

Amount of sales tax would be

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3 years ago

X^2-9x=x-9 solve by factoring

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(x-1)(x-9)
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2 years ago

Line segment CD is shown on a coordinate grid:

Lady_Fox [76]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

What is the equation for the plane illustrated below?

TiliK225 [7]

Answer:

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

Step-by-step explanation:

The general equation in rectangular form for a 3-dimension plane is represented by:

$a\cdot x + b\cdot y + c\cdot z = d$

Where:

$x$ , $y$ , $z$ - Orthogonal inputs.

$a$ , $b$ , $c$ , $d$ - Plane constants.

The plane presented in the figure contains the following three points: (2, 0, 0), (0, 2, 0), (0, 0, 3)

For the determination of the resultant equation, three equations of line in three distinct planes orthogonal to each other. That is, expressions for the xy, yz and xz-planes with the resource of the general equation of the line:

xy-plane (2, 0, 0) and (0, 2, 0)

$y = m\cdot x + b$

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - x-Intercept, dimensionless.

If $x_{1} = 2$ , $y_{1} = 0$ , $x_{2} = 0$ and $y_{2} = 2$ , then:

Slope

$m = \frac{2-0}{0-2}$

$m = -1$

x-Intercept

$b = y_{1} - m\cdot x_{1}$

$b = 0 -(-1)\cdot (2)$

$b = 2$

The equation of the line in the xy-plane is $y = -x+2$ or $x + y = 2$ , which is equivalent to $3\cdot x + 3\cdot y = 6$ .

yz-plane (0, 2, 0) and (0, 0, 3)

$z = m\cdot y + b$

$m = \frac{z_{2}-z_{1}}{y_{2}-y_{1}}$

Where:

$m$ - Slope, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - y-Intercept, dimensionless.

If $y_{1} = 2$ , $z_{1} = 0$ , $y_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

y-Intercept

$b = z_{1} - m\cdot y_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the yz-plane is $z = -\frac{3}{2}\cdot y+3$ or $3\cdot y + 2\cdot z = 6$ .

xz-plane (2, 0, 0) and (0, 0, 3)

$z = m\cdot x + b$

$m = \frac{z_{2}-z_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - z-Intercept, dimensionless.

If $x_{1} = 2$ , $z_{1} = 0$ , $x_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

x-Intercept

$b = z_{1} - m\cdot x_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the xz-plane is $z = -\frac{3}{2}\cdot x+3$ or $3\cdot x + 2\cdot z = 6$

After comparing each equation of the line to the definition of the equation of the plane, the following coefficients are obtained:

$a = 3$ , $b = 3$ , $c = 2$ , $d = 6$

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

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3 years ago

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