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2 years ago

Using the standard 28/36 ratio, determine the maximum allowable recurring debt for someone with a

Mathematics

1 answer:

Mumz [18]2 years ago

4 0

The maximum allowable recurring debt for someone with a monthly income of $54.875 is $4.39.

<h3 /><h3>Maximum allowable recurring debt:</h3>

Using this formula

Maximum allowable recurring debt=Ratio×Monthly income

Where:

Ratio=28/36

Monthly income=$54.875

Let plug in the formula

Maximum allowable recurring debt=(36%×$54.875)-(28%×$54.875)

Maximum allowable recurring debt=$19.755-$15.365

Maximum allowable recurring debt=$4.39

Inconclusion the maximum allowable recurring debt for someone with a monthly income of $54.875 is $4.39.

Learn more about maximum allowable recurring debt here:brainly.com/question/5083803

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What is the equation of a circle whose center is begin ordered pair negative 4 comma 6 end ordered pair and whose radius is 9 cm

solong [7]

Answer:

$(x+4)^2+(y-6)^2=81$

Step-by-step explanation:

We want to find equation of a circle with center (-4,6) and radius 9cm.

The equation of a circle with center (h,k) and radius $r$ units is given by;

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

We substitute the radius and the center to obtain;

$(x+4)^2+(y-6)^2=9^2$

The required equation is:

$(x+4)^2+(y-6)^2=81$

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3 years ago

Endpoints of segment MN have coordinates (0, 0), (5, 1). The endpoints of segment AB have coordinates (1 1/22 , 2 1/4 ) and (−2

Nana76 [90]

Answer: $k=18\dfrac{8}{11}$ .

Step-by-step explanation:

If a line passing through two points, then

$Slope=\dfrac{y_2-y_1}{x_2-x_1}$

Endpoints of segment MN have coordinates (0, 0) and (5, 1).

Slope of MN $=\dfrac{1-0}{5-0}=\dfrac{1}{5}$

The endpoints of segment AB have coordinates $\left(1\dfrac{1}{22} , 2\dfrac{1}{4}\right)$ and $\left(-2\dfrac{1}{4} , k\right)$ .

$A=\left(1\dfrac{1}{22} , 2\dfrac{1}{4}\right)=\left(\dfrac{23}{22} ,\dfrac{9}{4}\right)$

$B=\left(-2\dfrac{1}{4} , k\right)=\left(-\dfrac{9}{4} , k\right)$ .

Slope of AB $=\dfrac{k-\frac{9}{4}}{-\frac{9}{4}-\dfrac{23}{22}}$

$=\dfrac{\frac{4k-9}{4}}{\frac{-99-46}{44}}$

$=\dfrac{4k-9}{4}\times \dfrac{44}{-145}$

$=4k-9\times \dfrac{11}{-145}$

$=\dfrac{44k-99}{-145}$

Product of slopes of two perpendicular segments is -1.

Slope of MN × Slope of AB = -1

$\dfrac{1}{5}\times \dfrac{44k-99}{-145}=-1$

$\dfrac{44k-99}{-725}=-1$

$44k-99=725$

$44k=725+99$

$k=\dfrac{824}{44}$

$k=\dfrac{206}{11}$

$k=18\dfrac{8}{11}$

Therefore, the value of k is $k=18\dfrac{8}{11}$ .

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