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katen-ka-za [31]

2 years ago

8

Tony addison obtained a 20-year, $45,000 mortgage loan from united savings with an interest rate of 8.5%. his monthly payment is

$390.60. after the first payment, what is the new principal? $41,675.34 $42,395.68 $45,578.87 $44,928.15

1 answer:

Anna007 [38]2 years ago

4 0

As given,

Loan amount is = $45000

Rate of interest = 8.5%

So, Tony's mortgage will attract an interest of:

$45000\times\frac{8.5}{100}$ = $3825 (this is yearly)

And for 1st month it will be = $\frac{3825}{12}$ = $318.75

As given, the first month's payment is $390.60 and this covers the interest Additional amount ($390.60 - $318.75 = $71.85) is a payment against the principle.

Hence, the new principle after the 1st month is $71.85 less than $45000

= 45000-71.85 = $44928.15

Hence, the last option $44928.15 is the correct answer.

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If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ?

KIM [24]

Answer:

The answer is

$sin(\theta)=-\frac{\sqrt{2}}{2}$

$tan(\theta)=-1$

Step-by-step explanation:

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

$sin^{2}(\theta)+cos^{2}(\theta)=1$

In this problem we have

$cos(\theta)=\frac{\sqrt{2}}{2}$

$\frac{3\pi}{2}$

so

The angle $\theta$ belong to the third or fourth quadrant

The value of $sin(\theta)$ is negative

Step 1

Find the value of $sin(\theta)$

Remember

$sin^{2}(\theta)+cos^{2}(\theta)=1$

we have

$cos(\theta)=\frac{\sqrt{2}}{2}$

substitute

$sin^{2}(\theta)+(\frac{\sqrt{2}}{2})^{2}=1$

$sin^{2}(\theta)=1-\frac{1}{2}$

$sin^{2}(\theta)=\frac{1}{2}$

$sin(\theta)=-\frac{\sqrt{2}}{2}$ ------> remember that the value is negative

Step 2

Find the value of $tan(\theta)$

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=-\frac{\sqrt{2}}{2}$

$cos(\theta)=\frac{\sqrt{2}}{2}$

substitute

$tan(\theta)=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$

$tan(\theta)=-1$

8 0

2 years ago

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HELP FAST PLEASE

kramer

Answer:

y = 14

Step-by-step explanation:

proportional relation: y = kx

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2 years ago

What are the zeros of the quadratic function f(x) = 6x^2 + 12x – 7?

strojnjashka [21]

Answer:

$x1=\frac{-2+\sqrt{26/3}}{2}$

$x2=\frac{-2-\sqrt{26/3} }{2}$

Step-by-step explanation:

To find the zeros of the quadratic function f(x)=6x^2 + 12x – 7 we need to factorize the polynomial.

To do so, we need to use the quadratic formula, which states that the solution to any equation of the form ax^2 + bx + c = 0 is:

$x=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

So, the first thing we're going to do is divide the whole function by 6:

6x^2 + 12x – 7 = 0 -> x^2 + 2x - 7/6

This step is optional, but it makes things quite easier.

Then we using the quadratic formula, where:

a=1, b= 2, c = -7/6.

Then:

$x=\frac{-2±\sqrt{2^{2}-4(1)(-7/6)}}{2}$

$x=\frac{-2±\sqrt{4 +14/3}}{2}$

$x=\frac{-2±\sqrt{26/3}}{2}$

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$x1=\frac{-2+\sqrt{26/3}}{2}$

$x2=\frac{-2-\sqrt{26/3}}{2}$

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2 years ago

489 to the nearest ten

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2 years ago

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