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3 years ago

Write the equation of the line passing through the point (3,5) with a slope of 2

Mathematics

1 answer:

Nady [450]3 years ago

7 0

Y - y1 = m(x - x1)
y - 5 = 2(x - 3)
y - 5 = 2x - 6
y = 2x - 1

answer: equation y = 2x - 1

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The back of jake's property is a creek. Jake would like to enclose a rectangular area, using the creek as one side and fencing f

kow [346]

Answer:

The maximum possible area of the pasture is 26,450 feet square.

Step-by-step explanation:

Let us assume the width of the rectangular area = m ft

and the length of the area = k ft

Also assume: ( one side with k units IS NOT FENCED)

So, the total perimeter of the fencing is:

2 m + k = 460

or, k = 460 - 2m ........ (1)

Now, AREA OF THE RECTANGULAR PORTION = Length x Width = k x m

Put k = 460 - 2m

We get: A = m (460 - 2m)

or, A = 460 m - 2m²

We need to find the maximum for the parabolic function A = 460 m - 2m²

The function has a maximum value as the quotient in front of x^2 is negative: -2 < 0

A (max) = $c-\frac{b^2}{4a}$ where a = -2, b = 460, c = 0

= $-\frac{(460)^2}{4(-2)} = 26,450$

A max = 26,450 sq ft.

The maximum possible area of the pasture is 26,450 feet square.

7 0

4 years ago

HELP 30 pts:( Evaluate 2A - 3B for A = 7 and B = -2. 20 9 8

ASHA 777 [7]

Answer:

Step-by-step explanation:

3 0

3 years ago

Write as a mixed number in lowest terms 7.25

Verizon [17]

7.25 is the same as $7 \frac{25}{100}$ written as a mixed number. However, .25 is $\frac{1}{4}$ of 100.

In this case, while rewriting your mixed number, 7.25 is the same as $7\frac{1}{4}$ , so that would be your answer.

7.25 ⇒ 7 1/4

5 0

4 years ago

Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).

Firdavs [7]

Answer:

A) Particular solution:

$2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

B) Homogeneous solution:

$y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))$

C) The most general solution is

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

Step-by-step explanation:

Given non homogeneous ODE is

$y''+4y'+5y=10x+e^{-x}---(1)$

To find homogeneous solution:

$D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)$

To find particular solution:

$y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\$

Substituting $y_{p},y'_{p},y''_{p}$ in (1)

$y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\$

Equating the coefficients

$5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\$