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3 years ago

choose the point-slope form and f the equation below that represents the line through the point (6,-3) and has a slope of 1/2

Mathematics

2 answers:

AveGali [126]3 years ago

6 0

Answer:

The point-slope form of this equation would be y + 3 = 1/2(x - 6)

Step-by-step explanation:

In order to find this, start with the base form of point-slope form.

y - y1 = m(x - x1)

Now input the slope for m and the point for (x1, y1)

y - -3 = 1/2(x - 6)

y + 3 = 1/2(x - 6)

Elina [12.6K]3 years ago

6 0

Answer:

y+3= 1/2(x-6)

Step-by-step explanation:

y-y₁ = m(x-x₁) is point-slope form of equation of line where m is slope and (x₁,y₁) is a point that passes through the line .

From question statement,we observe that

Slope and a point is given.

Hence, m = 1/2 and (x₁,y₁) = (6,-3)

Putting above values in formula,we get

y - (-3) = 1/2(x-6)

y+3= 1/2(x-6) is point-slope form of equation that passes through a point (6,-3) and has a slope equals to 1/2.

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A cylinder has a volume of 216π cubic feet and a height of 18 feet. what is the area of the base? cylinder v = bh 1. substitute

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The cylinder is a three-dimensional shape, and the base area of the cylinder is 12π square feet

<h3>How to determine the base area?</h3>

The given parameters are:

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So, we have:

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Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

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VXYZ is a rectangle with a length of 20cm and a width of 12 cm. The length of the rectangle is increased by 30%. The width of th

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Step-by-step explanation:

the area is

length × width

increasing the length by 30% means multiplying the original length by 1.3.

increasing the width by 10% means multiplying the original width by 1.1.

so, the new area is (by using the old length and width)

length × 1.3 × width × 1.1 = length × width × 1.3×1.1 =

= old area × 1.3 × 1.1 = old area × 1.43

so, the area increases by 43%.

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