The graph of f(x) = x2 has been shifted into the form f(x) = (x − h)2 + k:

Solve -3+12-5(2y-7) The solution y is:

jolli1 [7]

Answer:

+12

Step-by-step explanation:

1. distribute the negative 5 to 2y and negative 7

2. combine like term's -3+12-10y+35

3.Get the Y value by itself -44 from both sides -44-2y=

-44 -44

4. divide negative 2 from both sides 2y=-44

-2= -2

8 0

3 years ago

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

F(x)=2x-3, solve when x=0,-1,2 answers A-3, -1,1 B-3,-5,1 C3,-1,1 D-3,-1,7

guajiro [1.7K]

Answer:

b

Step-by-step explanation:

8 0

2 years ago

F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0

2 years ago

A lawyer charges his clients a consultation fee of $500 and $100 per hour to work on a case. This is represented by the linear f

Elis [28]

They would make 2,000 because 150X10=1,500+500

5 0

2 years ago

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