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allsm [11]
2 years ago
5

Subtract 2y - 3 from 8y-2.

Mathematics
1 answer:
ivolga24 [154]2 years ago
3 0

Answer:

2(3y-2)

Step-by-step explanation:

Collect like terms

6y-4

Simple terms

2(3y-2)

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A certain college graduate borrows 7864 dollars to buy a car. The lender charges interest at an annual rate of 13%. Assuming tha
White raven [17]

Answer:

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = =$1573.17

Step-by-step explanation:

Consider A represent the balance at time t.

A(0)=$ 7864.

r=13 % =0.13

Rate payment = $k

The balance rate increases by interest (product of interest rate and current balance) and payment rate.

\frac{dB}{dt} = rB-k

\Rightarrow \frac{dB}{dt} - rB=-k.......(1)

To solve the equation ,we have to find out the integrating factor.

Here p(t)= the coefficient of B =-r

The integrating factor =e^{\int p(t) dt

                                     =e^{\int (-r)dt

                                     =e^{-rt}

Multiplying the integrating factor the both sides of equation (1)

e^{-rt}\frac{dB}{dt} -e^{-rt}rB=-ke^{-rt}

\Rightarrow  e^{-rt}dB - e^{-rt}rBdt=-ke^{-rt}dt

Integrating both sides

\Rightarrow \int e^{-rt}dB -\int e^{-rt}rBdt=\int-ke^{-rt}dt

\Rightarrow e^{-rt}B=\frac{-ke^{-rt}}{-r} +C        [ where C arbitrary constant]

\Rightarrow B(t)=\frac{k}{r} +Ce^{rt}

Initial condition B=7864 when t =0

\therefore 7864= \frac{k}{r} - Ce^0

\Rightarrow  C= \frac{k}{r} -7864

Then the general solution is

B(t)=\frac{k}{r}-( \frac{k}{r}-7864)e^{rt}

To determine the payment rate, we have to put the value of B(3), r and t in the general solution.

Here B(3)=0, r=0.13 and t=3

B(3)=0=\frac{k}{0.13}-( \frac{k}{0.13}-7864)e^{0.13\times 3}

\Rightarrow- 0.48\frac{k}{0.13} +11614.98=0

⇒k≈3145.72

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = ${(3145.72×3)-7864}

                                                 =$1573.17

4 0
3 years ago
Statistics urgent<br> Use n=6 and p=0.85 to complete
Bas_tet [7]

Answer:

I cant see anything on that picture. Wish I could help bruv. I wish I could help. It's a shame innit.

7 0
2 years ago
X) sec’A + cosec?A<br>secA. cosecA​
VikaD [51]

Answer:

(Sin A + Cos A)/Sin A. Cos A

Step-by-step explanation:

As we know

Sec A = 1/Cos A

and Cosec A = 1/Sin A

Given Equation

Sec A + Cosec A

Substituting the given values, we get -

1/cos A + 1/Sin A

(Sin A + Cos A)/Sin A. Cos A

3 0
2 years ago
For fun question
ZanzabumX [31]
Consider the function f(x)=x^{1/3}, which has derivative f'(x)=\dfrac13x^{-2/3}.

The linear approximation of f(x) for some value x within a neighborhood of x=c is given by

f(x)\approx f'(c)(x-c)+f(c)

Let c=64. Then (63.97)^{1/3} can be estimated to be

f(63.97)\approxf'(64)(63.97-64)+f(64)
\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375

Since f'(x)>0 for x>0, it follows that f(x) must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function f(x). This means the estimated value is an overestimation.

Indeed, the actual value is closer to the number 3.999374902...
4 0
3 years ago
Mrs. Siebenaller bought a bus for 25,000 with a 7% interest rate mrs s gets a loan payoff of 60 months how much interest would s
Artist 52 [7]

Answer:

\$4701.80

Step-by-step explanation:

Mrs. Siebenaller bought a bus for 25,000 with a 7% interest rate and she gets a loan payoff of 60 months,

We know that,

\text{PV of annuity}=P\left[\dfrac{1-(1+r)^{-n}}{r}\right]

Where,

PV = Present value of annuity = 25000,

r = rate of interest of each period = \dfrac{7}{12}% monthly

n = number of periods = 60 months,

Putting the values,

\Rightarrow 25000=P\left[\dfrac{1-(1+\frac{0.07}{12})^{-60}}{\frac{0.07}{12}}\right]

\Rightarrow P=\dfrac{25000}{\left[\dfrac{1-(1+\frac{0.07}{12})^{-60}}{\frac{0.07}{12}}\right]}

\Rightarrow P=\$495.03

Hence total amount paid is,

=495.03\times 60=\$29,701.80

Therefore interest amount is,

=29,701.80-25,000=\$4701.80


4 0
3 years ago
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