Hi I need HELP with this question

Mathematics

1 answer:

myrzilka [38]3 years ago

8 0

Answer:

Step-by-step explanation:

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TIMED TEST PLEASE HELP

xxMikexx [17]

1 ) first offer
total payments
375.76×12×4
=18,036.48
Interest paid
18,036.48−16,000
=2,036.48
Second offer
Total payments
390.61×12×4
=18,749.28
Interest paid
18,749.28−16,000
=2,749.28
Larry will save of taking 6% loan
2,749.28−2,036.48=712.8. .answer

2) credit card 1
Total payments
277.09×12
=3,325.08
Interest paid
3,325.08−3,000
=325.08
Credit card 2
Total payments
152.69×12×2
=3,664.56
Interest paid
3,664.56−3,000
=664.56
Susan will save
664.56−325.08
=339.48...answer

Hope it helps!

6 0

3 years ago

Read 2 more answers

0.5x=0.3x-9 what is x

sleet_krkn [62]

O.5 0.3 = x= 0.8 should be

3 0

3 years ago

Read 2 more answers

Identify the standard form of the equation by completing the square.

OLEGan [10]

Answer:

$\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Step-by-step explanation:

<u>Given equation</u>:

$4x^2-9y^2-8x+36y-68=0$

This is an equation for a horizontal hyperbola.

<u>To complete the square for a hyperbola</u>

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

$\implies 4x^2-8x-9y^2+36y=68$

Factor out the coefficient of the x² term and the y² term.

$\implies 4(x^2-2x)-9(y^2-4y)=68$

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

$\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2$

$\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36$

Factor the two perfect trinomials on the left side:

$\implies 4(x-1)^2-9(y-2)^2=36$

Divide both sides by the number of the right side so the right side equals 1:

$\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}$

Simplify:

$\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Therefore, this is the standard equation for a horizontal hyperbola with:

center = (1, 2)
vertices = (-2, 2) and (4, 2)
co-vertices = (1, 0) and (1, 4)
$\textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}$
$\textsf{Foci}: \quad (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)$