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3 years ago

Can someone please help me solve this problem????

2 answers:

7 0

For this problem, we will be using the formula for loan:

PMT=P[(r/n)/1-(1+r/n)^-ny]
where:
P=Principal Value
r=rate
n=number of compoundings/year
y=year

To solve:
*Weston is only financing $185,000.

P=$185,000
r=6.525% or 0.06525
n=12 (monthly)
year=30

PMT=185000[(0.06525/12)/1-(1+0.6525/12)^-12*30
= 185000[(.0054375)/1-(1+.0054375)^-360
= 185000[(.0054375)/1-(1.0054375)^-360
= 185000[(.0054375)/1-(0.142)
= 185000[(.0054375)/(.858)
= 185000(0.00634)
= 1,172.37

Answer: $1,172.37

vodka [1.7K]3 years ago

3 0

$1,172.37 The formula for calculating the payment on a loan is P = r(PV)/(1 - (1 + r)^(-n)) where P = Payment PV = Present value r = interest per period n = number of periods Since there's 12 months per year and the loan is for 30 years, there will be 12 * 30 = 360 periods. The interest rate per month will be 0.06525 / 12 = 0.0054375, and finally, the present value is the size of the loan at 185000. Now let's substitute and solve. P = r(PV)/(1 - (1 + r)^(-n)) P = 0.0054375(185000)/(1 - (1 + 0.0054375)^(-360)) P = 1005.9375/(1 - (1.0054375)^(-360)) P = 1005.9375/(1 - 0.141961802) P = 1005.9375/0.858038198 P = 1172.369134 P = 1172.37 So the monthly interest and principle payments are $1,172.37 Note: The actual payments will be higher since the above figure doesn't include insurance and taxes.

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2x3 + 5x2 + 6x + 15

iren [92.7K]

Answer:

Step-by-step explanation:

Distribute x^2 to 2x and 5, you’ll get 2x^3+5x^2.

The distribute 3 to 2x and 5, you’ll get 6x+15.

You’ll get what the question is asking for.

7 0

3 years ago

What is the area of the shaded region below? 7 m 30 m 7 m 21 m

Rama09 [41]

Answer:

322.28

Step-by-step explanation:

Area of the rectangle

L = 30

W = 21

Area = 30 * 21

Area = 630

Area of the circles

Radius = 7 m

Area = pi r^2

Area = 3.14 * 7^2

Area = 153.86 m^2

Area of both circles = 307.72

Area of the Shaded Region

Area of shaded region = area of the rectangle - area of 2 circles.

Area of Shaded Region = 630 - 307.72

Area of Shaded Region = 322.28

5 0

2 years ago

in a marathon race (42.195 km), the winner’s time is 2 hours, 13 minutes, and 9 seconds, while the second-place time is 2 hours,

algol [13]

in a marathon race (42.195 km), the winner’s time is 2 hours, 13 minutes, and 9 seconds, while the second-place time is 2 hours, 26 minutes, and 6 seconds . The distance is 2.664 km.

D = distance of marathon = 42.195 km = 42195 m

tw = time taken by winner = 2 h 17 min = 2 (60) + 17 = 137 min = 137 (60) = 8220 sec

Vw = speed of winner

speed of winner is given as

Vw = D /tw = 42195/8220

ts = time taken by second place = 2 h 26 min 14 sec = 2 (60)(60) + (26)(60) + 14 = 8774 s

Vs = speed of second place

speed of second place is given as

Vs = D /ts = 42195/8774

distance of the second place holder from the finish line is given as

D' = Vs (ts - tw)

D' = (42195/8774 )(8774 - 8220)

D' = 2664.24 m

D'= 2.664 km

To learn more about distance visit:brainly.com/question/15172156

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7 0

1 year ago

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means: