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mr_godi [17]
3 years ago
15

The​ Adeeva's gross monthly income is ​$5600. They have 18 remaining payments of $ 360 on a new car. They are applying for a 15​

-year, ​$142 comma 000 mortgage at 7.5​%. The taxes and insurance on the house are ​$270 per month. The bank will only approve a loan that has a total monthly mortgage payment of​ principal, interest, property​ taxes, and​ homeowners' insurance that is less than or equal to​ 28% of their adjusted monthly income. What is 28% of Adevas income
Mathematics
1 answer:
olga55 [171]3 years ago
5 0

Answer:

$1,467.20

Step-by-step explanation:

Adjusted monthly income is the gross income minus any payments made every month. 5600 - 360 = 5240. To get 28% of her adjusted income multiply 5240 X.28 =1467.20

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If sin square theta = 0.75, then cos =
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Answer:

\cos\theta=\pm0.5

Step-by-step explanation:

Given:

\sin^2\theta=0.75

To find \cos\theta

Using trigonometric relations for sums and differences of squares of the ratios.

We know:

\sin^2\theta+\cos^2\theta =1

Plugging in  \sin^2\theta=0.75 in the above relation.

0.75+\cos^2\theta =1

Subtracting both sides by 0.75.

0.75+\cos^2\theta-0.75 =1-0.75

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\sqrt{\cos^2\theta} =\sqrt{0.25}

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If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle? A) (x
Andrei [34K]

Answer:

\large\boxed{A.\ (x+6)^2+(y+10)^2=20}

Step-by-step explanation:

The standard form of an equation of a circle:

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

(h, k) - center

r - radius

We have the endpoints of the diameter of a circle (-8, -6) and (-4, -14).

The midpoint of a diameter is a center of a circle.

The formula of a midpoint:

\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)

Substitute:

x=\dfrac{-8+(-4)}{2}=\dfrac{-12}{2}=-6\\\\y=\dfrac{-6+(-14)}{2}=\dfrac{-20}{2}=-10

We have h = -6 and k = -10.

The radius is the distance between a center and the point on a circumference of a circle.

The formula of a distance between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Substitute (-6, -10) and (-8, -6):

r=\sqrt{(-8-(-6))^2+(-6-(-10))^2}=\sqrt{(-2)^2+4^2}=\sqrt{4+16}=\sqrt{20}

Finally we have

(x-(-6))^2+(y-(-10))^2=(\sqrt{20})^2\\\\(x+6)^2+(y+10)^2=20

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4 years ago
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